FBISE Physics Numerical Problems

This Page is dedicated for the Numerical Problems of Physics for the Federal Board of intermediate and Secondary Education (FBISE), Islamabad, Pakistan

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FBISE Physics MCQs for Class 9

Prepare for your FBISE Physics exam of Class 9 with our MCQs practice and tips. Learn the best strategies for answering multiple choice questions and boost your scores!

BISE Physics MCQs for Class 9 of Unit 1 Physical Quantities and Measurements

1.1 Express the following quantities using prefixes.

(a) 5000 g (b) 2000 000 W (c) 52 x10^-10 kg (d) 225x10^-8s

Solution

(a) 5000 g = 5x10³g = 5 kg

(b) 2000 000 W = 2x10⁶W = 2 MW

(d) 225x10^-8s = 2.25x10²x10^-8s = 2.25x10^-6s = 2.25 µg

1.2 How do the prefixes micro, nano and pico relate to each other?

Solution

micro = 10^-6

nano= 10^-9

pico= 10^-12

i. Relation between micro and nano

micro= 10^-6= 1x10^-6=10³x10^-3x10^-6= 10³x 10^-9 = 10³nano.

ii. Relation between micro and pico

micro= 10^-6= 1x10^-6=10⁶x10^-6x10^-6= 10⁶x 10^-12 = 10⁶pico.

iii. Relation between nano and micro

nano= 10^-9= 1x10^-9=10^-3x10^-6= 10^-3micro.

iv. Relation between nano and pico

nano = 10^-9= 1x10^-9=10³x10^-3x10^-9= 10³x 10^-12 = 10³pico.

v. Relation between pico and micro

pico= 10^-12=10^-6x10^-6= 10^-6micro.

vi. Relation between pico and nano

pico= 10^-12= 10^-3x10^-9= 10^-3nano.

1.3 Your hair grows at the rate of 1 mm per day. Find their growth rate in nm s^-1.

Data

Hair growth (mm/day) = 1

Hair growth (nms^-1) =?

Solution

Hair growth = 1 mm/day = 1x10^-3 m/day

= 1x10-3 m/ (24x60x60 s)

= 1x10^-3 m/86400 s

= 1.1574x10^-8ms^-1

= 11.574 x10^-1x10^-8 ms^-1

= 11.574x10^-9ms^-1

Hair growth in nms^-1 = 11.574 nms^-1Ans

1.4 Rewrite the following in standard form.

(a) 1168x10^-27 (b) 32x10^-5 (c) 725 x10^-5 kg (d) 0.02 x10^-8

(a) 1168x10^-27

Solution

1168x10^-27= 1.168x10³x10^-27= 1.168x10^-27+3= 1.168x10^-24

(b) 32x10^-5

Solution

32x10^-5= 3.2x10¹x10^-5= 32x10^-5+1= 3.2x10^-4

(c) 725 x10^-5 kg

Solution

725 x10^-5 kg = 7.25x10² x10^-5 kg

= 7.25 x10^-5+2 kg

= 7.25 x10^-3 x10³ g

= 7.25 x10^-3+3 g

= 7.25 x10⁰ g

= 7.25 x 1

= 7.25 g

(d) 0.02 x10^-8

Solution

0.02 x10^-8= 2x10^-2 x10^-8= 2 x10^-8-2= 2 x10^-10= 0.02 x10^-8= 2 x10^-8

Write the following quantities in standard form.
(a) 6400 km
(b) 380 000 km
Solution

1.5 (c) 300 000 000 ms-1
(d) seconds in a day

(a) 6400 km

Solution

6400 km = 6.4x10³ km

(b) 380 000 km

Solution

380 000 km = 3.8x10⁵km

(c) 300 000 000 ms^-1

Solution

300 000 000 ms^-1= 3x10⁸ms^-1

(d) seconds in a day

Solution

seconds in a day = 24x60x60 s = 86400 s = 8.64x10⁴s

1.6 On closing the jaws of a Vernier Callipers, zero of the vernier scale is on the right to its main scale such that 4th division of its vernier scale coincides with one of the main scale division. Find its zero error and zero correction.

Solution

Main scale reading = 0.0 cm

Vernier division coinciding with main scale = 4 div.

Vernier scale reading = (Vernier division coinciding with main scale) (Least Count of vernier Calliper)

= 4 x 0.01 cm = 0.04 cm

Zero error = (Main Scale reading) + (Vernier Scale Reading)

= 0.0 cm + 0.04 cm

As zero line of vernier scale is on the right of main scale zero, therefore zero error is positive.

Zero error = + 0.04 cm

zero correction (Z.C) = - 0.04 cm

1.7 A screw gauge has 50 divisions on its circular scale. The pitch of the screw gauge is 0.5 mm. What is its least count?

Solution

Divisions on Circular Scale = 50

Pitch of Screw Gauge = 0.5 mm

Least Count of Screw Gauge =?

\[Least Count=\frac{Pitch of Screw gauge}{Number of divisions on circular scale}\]

\[Least Count=\frac{0.5 mm}{50}\]

\[Least Count=0.01 mm\]

1.8 Which of the following quantities have three significant figures?

(a) 3.0066 m (b) 0.00309 kg (c) 5.05x10^-27 kg (d) 301.0 s

(a) 3.0066 m

Solution

All the 5 digits are significant.

(b) 0.00309 m

Solution

The first two zeros are not significant. They are used to space the decimal point. The digits 3 and 9 are significant. The zero between the two significant figures 3 and 9 are significant. Thus, there are three significant figures. In scientific notation this number can be written

0.00309 m = 3.09x10^-2 m

(c) 5.05x10^-27 kg

Solution

Both the 5 are significant. The zero between the two significant figures 5 and 5 are significant. Thus, there are three significant figures

(d) 301.0 s

Solution

The final zero is significant since it comes after the decimal point. The zero between 3 and1 is also significant because it comes between the significant figures. Thus, the number of significant figures in this case is four. In scientific notation, it can be written as
301.0 s = 3.010x10² s

1.9 What are the significant figures in the following measurements?

(a) 1.009 m (b) 0.00450 kg (c) 1.66x10^-27 kg (d) 2001 s

(a) 1.009 m

Solution

The digits 1 and 9 are significant. The zeros between the two significant figures 1 and 9 are also significant.

(b) 0.00450 kg

Solution

The first two zeros are not significant. They are used to space the decimal point. The digits 4 and 5 are significant. The final or ending zero is also significant. Thus, there are three significant figures. In scientific notation this number can be written

0.00450 m = 4.50x10^-3 m

(c) 1.66x10^-27 kg

Solution

All the three (03) digits are significant.

(d) 2001 s

Solution

All the four digits are significant.

Solution

1.10 A chocolate wrapper is 6.7 cm long and 5.4 cm wide. Calculate its area up to reasonable number of significant figures.

Solution

Length of Chocolate wrapper = 6.7 cm

Width of chocolate wrapper = 5.4 cm

Area of chocolate wrapper =?

Area of chocolate wrapper = (Length of chocolate wrapper) (width of chocolate wrapper)

Area of chocolate wrapper = 6.7 cm x 5.4 cm

Area of chocolate wrapper = 36.18 cm²

FBISE Physics MCQs for Class 9 of Unit 2 Kinematics

2.1 A train moves with a uniform velocity of 36 kmh^-1 for 10 s. Find the distance travelled by it.

Solution

Data

velocity of train = v = 36 kmh^-1

= 36 km/h

= 36x1000/3600 m/s

= 10 m/s

Time = t = 10 s

Distance travelled by the train = S = ?

Formula

S = vt

Solution

S=vt

Putting values in this equation

S = (10 m/s) (10 s)

S = 100 m Ans

2.2 A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. What will be its speed at the end of 100 s.?

Solution

Data

Initial speed of the train = v_i = 0 (Because initially the train is at rest)

Distance travelled by the train = S = 1 km = 1000 m

Time = t = 100 s

Speed of train at the end of 100 s = Final speed of the train = v_f= ?

Formula

\[S=v_{i}t + \frac{1}{2}at^{2} \]

\[v_{f}=v_{i} + at\]

Solution

First we wiil calculate the acceleration of the train using 2nd equation of motion i.e.

\[S=v_{i}t + \frac{1}{2}at^{2} \]

Putting values in this equation

\[1000 =(0))(100) + \frac{1}{2}(a)(100))^{2}\]

\[1000 = 0 +\frac{1}{2}(a)(10000)\]

\[1000 = \frac{1}{2}(a)(10000)\]

\[2000 = (a)(10000)\]

\[\frac{2000}{10000}= a\]

\[a = 0.2 ms^{-2}\]

Putting value of acceleration a in 1st equation of motion i.e.

\[v_{f}=v_{i} + at\]

\[v_{f}= 0 + (0.2 ms^{-2}))(100 s))\]

\[v_{f}= (0.2 ms^{-2}))(100 s))\]

\[v_{f}= (20 ms^{-1})\]

v_f = 20 ms^-1Ans

The velocity of train at the end of 100 s is v_f = 20 ms^-1

2.3 A car has a velocity of 10 ms^-1. It accelerates at 0.2 ms^-2 for half minute. Find the distance travelled during this time and the final velocity of the car.

Solution

Data

Initial velocity of the car= v_i = 10 ms^-1

Acceleration of the car = a = 0.2 ms^-2

Time of acceleration = t = half minute = 30 s

Distance travelled by the car in 30 s = S =?

Final velocity of the car = v_f=?

Formula

\[v_{f}=v_{i} + at\]

\[S=v_{i}t + \frac{1}{2}at^{2} \]

Solution

First, we will calculate the final velocity of the car using first equation of the motion i.e.

\[v_{f}=v_{i} + at\]

Putting values in this equation

\[v_{f}= (10 ms^{-1} + (0.2 ms^{-2})(30 s)\]

\[v_{f}= (10 ms^{-1} + 6 ms^{-1}\]

\[v_{f}= 16 ms^{-1}\]

Final velocity of the car is v_{f =}16 ms^-1

Now we will calculate the calculate the distance travelled by the car in 30s using 2^nd equation of motion i.e.

\[S=v_{i}t + \frac{1}{2}at^{2} \]

Putting values in this equation

\[S =(10 ms^{-1})(30 s) + \frac{1}{2}(0.2 ms^{-2})(30))^{2}\]

\[S = 300 m + \frac{1}{2}(0.2 ms^{-2})(900 s_{2})\]

\[S = 300 m + \frac{1}{2}(180 m)\]

\[S = 300 m + 90 m\]

\[S = 390 m\]

The distance travelled by the car after 30 s is 390 m.

2.4 A tennis ball is hit vertically upward with a velocity of 30 ms^-1. It takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return to ground?

Solution

Data

Initial velocity of the ball = v_i = 30 ms^-1

Time taken by the ball to reach the highest point = t= 3 s

Final velocity of the ball = the velocity at the highest point = v_f= 0

Maximum height reached by the ball = S =?

Time taken by the ball to return the ground = T =?

Acceleration of the ball = a = -g = -10 ms^-1

Formula

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

First, we will calculate the maximum height reached by the ball using 3^rd equation of motion i.e.

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation

\[2aS=v_{f}^{2}-v_{i}^{2}\]

\[2(-10))S= 0^{2} - 30^{2}\]

\[-20S= 0 - 900\]

\[-20S=- 900\]

\[S= 45 m\]

Now, we will find the time taken by the ball to reach the ground

Time taken by the ball to return the ground

\[\left ( Time taken by the ball to retur to groun \right ) =\left (Time taken by the ball to reach the highest point\right ) + \left (Time taken by the ball from highest point to ground \right )\]

Time taken by the ball to return to ground = 3 s + 3 s

Time taken by the ball to return to ground = 6 s Ans

2.5 A car moves with uniform velocity of 40 ms^-1 for 5 s. It comes to rest in the next 10 s with uniform deceleration. Find

(i) deceleration

(ii) total distance travelled by the car.

Solution

Data

Initial velocity of the car = v_i = 40 ms^-1

Time for which car travels at the rate of 40 ms^-1= t₁ = 5 s

Time taken by the car for coming to rest from the speed of 40 ms^-1 = t₂ = 10 s

Deceleration of the car = a =?

Total distance travelled by the car =?

Formula

Part I

Initial velocity of the car = v_i = 40 ms^-1

Final velocity of the car = v_f= 0

Time taken by the car for coming to rest from the speed of 40 ms^-1 = t₂ = 10 s

Deceleration of the car = a =?

Formula

\[a= \frac{v_{f}- v_{i}}{t}\]

Putting values in this equation

\[a= \frac{0 - 40}{10}\]

\[a= \frac{- 40}{10}\]

\[a= - 4 ms^{-1} Ans\]

Part II

We will calculate the distance in three steps.

In 1^st step we will calculate the distance S₁ for first 5 s when the velocity is unform.

In 2^nd step we will calculate the distance S₂ for 10 s when the velocity is decreasing.

In 3^rd step we will add up these distances S₁and S₂to get the total distance S

Step I

velocity = Initial velocity of the car = v_i = 40 ms^-1

tine = Time for which car travels at the rate of 40 ms^-1= t₁ = 5 s

Distance travelled by the car in initial 5 s = S₁ =?

Formula

S=vt

Solution

S₁ = v_it₁

putting values in this equation

S₁ = (40)(5)

S₁ = 200 ms^-1

Step II

Initial velocity of the car = v_i = 40 ms^-1

Final velocity of the car = v_f= 0

Deceleration of the car = a = - 4 ms^-2

Distance for 10 s when the velocity is decreasing = S₂= ?

Formula

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

\[2aS_{2}=v_{f}^{2}-v_{i}^{2}\]

Rearrange this equation for S₂

\[S_{2}=\frac{(0)^{2}- (40)^{2}}{2(-4)}\]

\[S_{2}=\frac{0 - 1600}{-8}\]

\[S_{2}=\frac{- 1600}{-8}\]

\[S_{2}= 200 m\]

Total distance travelled by the car = S = S₁+ S₂

Total distance travelled by the car = S = 200 m + 200 m

Total distance travelled by the car = S = 400 m Ans

2.6 A train starts from rest with an acceleration of 0.5 ms^-1. Find its speed in kmh^-1 when it has moved through 100 m.

Solution

Data

Initial speed of the train = v_i = 0 ms^-1 (initially the train is at rest)

Acceleration of the train = a = 0.5 ms^-1

Distance travelled by the train = 100 m

Speed of the train after moving a distance of 100 m = Final speed of the train = v_f = ?

Formula

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation

\[2aS=v_{f}^{2}-v_{i}^{2}\]

\[2(0.5))(100) =v_{f}^{2}-0^{2}\]

\[100 = v_{f}^{2}- 0 \]

\[v_{f}^{2} = 100 \]

\[v_{f} = 10 ms^{-1} \]

Now we will convert this speed into kmh^-1.

\[v_{f} = \frac{(10)(3600)}{1000} kmh^{-1} \]

\[v_{f} = 36 kmh^{-1} \]

Speed of the train after travelling a distance of 100 m is 36 kmh^-1

2.7 A train staring from rest, accelerates uniformly and attains a velocity 48 kmh^-1 in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes. Find the total distance travelled by the train.

Solution

Data

Initial speed of the train = v_i = 0 ms^-1 (initially the train is at rest)

Time for which train accelerates uniformly = t₁ = 2 minutes = 120 s

Speed of train after 2 minutes = 48 kmh^-1 = 13.33 ms^-1

Time for which train travels with speed of 48 kmh^-1 = t₂ = 5 minutes = 300 s

Time for uniform retardation of train = t₃ = 3 minutes = 180 s

Total distance travelled by the train = S =?

Formula

We will divide this problem into three parts and calculate the distance travelled by the train in each part separately and then add these distances to get the total distance.

Part I

Initial speed of the train = v_i = 0 ms^-

final speed of the train = v_f = 13.33 ms^-1

Time = 120 s

Distance travelled by train during this time = S₁= ?

Formula

\[v_{f}=v_{i} + at\]

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

First, we will calculate the acceleration of the of the train for this part using 1^st equation of motion

\[v_{f}=v_{i} + at\]

Solving this equation for acceleration a

\[a=\frac{v_{f}-v_{i}}{t}\]

Putting values in this equation

\[a=\frac{13.33 - 0}{120}\]

\[a=\frac{13.33}{120}\]

\[a= 0.111 ms^{-1}\]

Now, we will calculate the distance S₁ travelled by the train using 3^rd equation of motion

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solving this equation for S

\[2aS_{1}=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation

\[S_{1}=\frac{v_{f}^{2}-v_{i}^{2}}{2a}\]

\[S_{1}==\frac{(13.33)^{2}-(0))^{2}}{2(0.111)}\]

\[S_{1}==\frac{177.6889- 0}{0.222)}\]

\[S_{1}==\frac{177.6889}{0.222)}\]

\[S_{1}== 800.4 m\]

Part II

velocity of the train = v = 13.33 ms^-1

Time of motion = t = 300 s

Distance travelled by the train during this time = S =?

Formula

S=vt

Solution

S₂=vt

S₂= (13.33) (300)

S₂ = 3999 m

Part III

Initial speed of the train = v_i = 13.33 ms^-1

final speed of the train = v_f = 0 ms^-1

Time = 180 s

Distance travelled by train during this time = S₂=?

Formula

\[v_{f}=v_{i} + at\]

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

First, we will calculate the acceleration of the of the train for this part using 1^st equation of motion

\[v_{f}=v_{i} + at\]

Solving this equation for acceleration a

\[a=\frac{v_{f}-v_{i}}{t}\]

Putting values in this equation

\[a=\frac{0 - 13,33}{180}\]

\[a=\frac{13.33}{180}\]

\[a= - 0.074 ms^{-1}\]

Now, we will calculate the distance S₃ travelled by the train using 3^rd equation of motion

\[2aS_{3}=v_{f}^{2}-v_{i}^{2}\]

Solving this equation for S₃

\[S_{3}=\frac{v_{f}^{2}-v_{i}^{2}}{2a}\]

Putting values in this

\[S_{3}==\frac{(0)^{2}-(13.33))^{2}}{2(-0.074)}\]

\[S_{3}==\frac{0 - 177.6889}{-0.148}\]

\[S_{3}==\frac{-177.6889}{-0.148)}\]

\[S_{3}== 1200.6 m\]

Total Distance = S = S₁+ S₂+ S₃

Total Distance = S = 800.4 m + 3999 m + 1200.6 m

Total Distance = S = 6000 m

2.8 A cricket ball is hit vertically upwards and returns to ground 6 s later. Calculate

(i) maximum height reached by the ball,

(ii) initial velocity of the ball.

Solution

Data

Time after which ball returns to the ground = T = 6 s

Acceleration of the ball for upward motion = a = -g = -10 ms^-1

velocity of the ball at the highest point = final velocity of the ball = v_f = 0

(when the ball reach at the highest point it comes to rest

Maximum height reached by the ball = S =?

Initial velocity of the ball = v_i= ?

Formula

\[v_{f}=v_{i} + at\]

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

First, we will calculate the initial velocity of the ball using 1^st equation of motion

\[v_{f}=v_{i} + at\]

We will consider only the upward motion

As 6s is the time of the whole round trip (i.e., time for both upward and downward motion),

therefore the time for upward motion is half of this time i.e. 3s

Therefore, here we will use t= 3s

Putting values in 1^st equation of motion

\[0 =v_{i} + (-10))(3)\]

\[0 =v_{i} - 30\]

\[v_{i} = 30 ms^{-1}\]

Now, we will calculate the maximum height reached by the ball

For this purpose, we will use 3^rd equation of motion

\[2aS=v_{f}^{2}-v_{i}^{2}\]

We will consider the upward motion only

Putting values in 3^rd equation of motion

\[2(-10)S= (0)^{2}-(30)^{2}\]

\[-20S= 0 - 900 \]

\[-20S= - 900 \]

\[S= 45 m \]

2.9 When brakes are applied, the speed of a train decreases from 96 kmh^-1 to 48 kmh^-1 in 800 m. How much further will the train move before coming to rest?

(Assuming the retardation to be constant).

Solution

Data

Initial Speed of the train = v_i = 96 kmh^-1= 26.67 ms^-1

Final Speed of the train = v_f= 48 kmh^-1= 13.33 ms^-1

Distance travelled by the train during decrease of speed from 96 kmh^-1to 48 kmh^-1= S₁ = 800 m

How much further will the train move before coming to rest = S₂=?

Formula

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

We will solve this problem in two parts, in first part we will calculate the retardation of the train

using 3^rd equation of motion and in second part we will calculate the distance travelled by the

train before coming to rest

Part I

Now, 1^st we will calculate the deceleration of the train using 3^rd equation of motion

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation

\[2a(800)= (13.33)^{2}-(26.67)^{2}\]

\[(1600)a= (13.33)^{2}-(26.67)^{2}\]

\[(1600)a= 177.6889 - 711.2889 \]

\[(1600)a= -533.6 \]

\[a= -0.3335 ms^{-2} \]

Now we will calculate the distance travelled by the train using 3^rd equation of motion

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation

\[2(-0.3335)S= (0)^{2}-(26.67)^{2}\]

\[(-0.667)S= 0 - 711.2889\]

\[(-0.667)S= 711.2889\]

\[S= 1066.4 m \]

Now

Total distance travelled by the train = distance travelled by the train during decrease of speed from 96 kmh^-1 to 96 kmh^-1 + Further distance travelled by the train before coming to rest

\[S= S_{1} + S_{2}\]

\[S_{2} = S - S_{1} \]

\[S_{2} = 1066.4 - 800 \]

\[S_{2} = 266.4 m \]

\[S_{2} = 266 m \]

2.10 In the above problem, find the time taken by the train to stop after the application of brakes.

Solution

Data

Initial Speed of the train = v_i = 96 kmh^-1= 26.67 ms^-1

Final Speed of the train = v_f= 48 kmh^-1= 13.33 ms^-1

Distance travelled by the train during decrease of speed from 96 kmh^-1to 48 kmh^-1= S₁ = 800 m

time taken by the train to stop after the application of brakes = t =?

Formula

\[v_{f}=v_{i} + at\]

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

We will solve this problem in two parts, in first part we will calculate the retardation of the train

using 3^rd equation of motion and in second part we will calculate the time taken by the train to

stop after the application of brakes.

Part I

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation

\[2a(800)= (13.33)^{2}-(26.67)^{2}\]

\[(1600)a= (13.33)^{2}-(26.67)^{2}\]

\[(1600)a= 177.6889 - 711.2889 \]

\[(1600)a= -533.6 \]

\[a= -0.3335 ms^{-2} \]

Part II

Now, we will calculate the time taken by the train to stop after the application of brakes.

When the train stops its velocity becomes zero. Therefore, our new data becomes

Initial Speed of the train = v_i = 96 kmh^-1= 26.67 ms^-1

Final Speed of the train = v_f= 0

time taken by the train to stop after the application of brakes = t =?

We will calculate the time taken by the train to stop after the application of brakes using 1^st

equation of motion.

\[v_{f}=v_{i} + at\]

We will re arrange this equation to get the value of

\[t= \frac{v^{f}-v^{i}}{a}\]

Putting values in this equation

\[t= \frac{0 - 26.67}{-0.3335}\]

\[t= \frac{- 26.67}{-0.3335}\]

\[t= 79.97 s\]

By rounding off the answer

t = 80 s Ans

FBISE Physics MCQs for Class 9 of Unit 3 Dynamics

3.1 A force of 20 N moves a body with an acceleration of 2 ms^-2. What is its mass?

Solution

Data

Force applied on body = F = 20 N

Acceleration produced in the body = a = 2 ms^-1

Mass of the body = m =?

Formula

sF = ma

Solution

F=ma

Rearranging this equation for m

\[m = \frac{F}{a} \]

Putting values in this equation

\[m = \frac{20}{2} \]

m = 10 kg Ans

3.2 The weight of a body is 147 N. What is its mass?

(Take the value of g as 10 ms^-2)

Solution

Data

Weight of the body = w = 147 N

Value of gravitational acceleration = g = 10 ms^-2

Mass of the body = m =?

Formula

w = mg

Solution

w=mg

Rearranging this equation for m

\[m = \frac{w}{g} \]

putting values in this equation

\[m = \frac{147}{10} \]

m = 14.7 kg Ans

3.3 How much force is needed to prevent a body of mass 10 kg from falling?

Solution

Data

Mass of the body = m = 10 kg

Value of gravitational acceleration = g = 10 ms^-2

Force required to prevent the body from falling = F =?

Formula

F = ma

Solution

F=ma

Here

a = g =10 ms^-2

Putting values in the above equation

F = (10)(10)

F = 100 N Ans

3.4 Find the acceleration produced by a force of 100 N in a mass of 50 kg.

Solution

Data

Force applied on mass = F = 100 N

Mass of the body = m = 50 kg

Acceleration produced in the body = a = ?

Formula

F = ma

Solution

F=ma

Rearranging this equation for acceleration a

\[a = \frac{F}{m} \]

\[m = \frac{100}{50} \]

a = 2 ms^-2 Ans

3.5 A body has weight 20 N. How much force is required to move it vertically upward with an acceleration of 2 ms^-2?

Solution

Data

Weight of the body = w = 20 N

Acceleration with which body moves upward = a = 2 ms^-2

Value of gravitational acceleration = g = 10 ms^-2

Force is required to move it vertically upward = F =?

Formula

w = mg

F = ma

Solution

Total Force required to move the body vertically upward = Reaction force + Force required produce acceleration

F = F_R+ F_a --------- (1)

First, we will calculate the mass of the body

w= mg

Re arranging this equation for m

\[m = \frac{w}{g} \]

\[m = \frac{20}{10} \]

m = 2 kg

Reaction force = F_R= w = mg = (2)(10) = 20 N

Force required produce acceleration = F_a = ma = (2)(2) = 4 N

Putting these values in equation (1)

F = F_R+ F_a

F = 20 N+ 2 N

F = 22 N Ans

3.6 Two masses 52 kg and 48 kg are attached to the ends of a string that passes over a frictionless pulley. Find the tension in the string and acceleration in the bodies when both the masses are moving vertically.

Solution

Data

Mass of 1^st body = m₁ = 52 kg

Mass of 2^nd body = m₂ = 48 kg

value of gravitational acceleration = g = 10 ms^-2

Tension in the string = T =?

Acceleration in the bodies = a =?

Formula

\[T = \frac{2m_{1}m_{2}}{m_{1}+m_{2}} \]

\[a = \frac{m_{1} - m_{2}}{m_{1}+m_{2}} \]

Solution

First, we will calculate the tension in the string

\[T = \frac{2m_{1}m_{2}}{m_{1}+m_{2}} \]

Putting values in this equation

\[T = \frac{2(52)(48))}{52 + 48} \]

\[T = \frac{4992}{100} \]

T = 499.2 N

Rounding the answer

T = 500 N Ans

Now, we will calculate the acceleration in the bodies

\[a = \frac{m_{1} - m_{2}}{m_{1}+m_{2}} \]

putting values in this equation

\[a = \frac{52 - 48}{52 + 48} \]

\[a = \frac{4}{100} \]

a = 0.04 ms^-2 Ans

3.7 Two masses 26 kg and 24 kg are attached to the ends of a string which passes over a frictionless pulley. 26 kg is lying over a smooth horizontal table. 24 kg mass is moving vertically downward. Find the tension in the string and the acceleration in the bodies.

Solution

Data

mass moving vertically downward = m₁ = 24 kg

mass lying over horizontal table = m₂ = 26 kg

value of gravitational acceleration = g = 10 ms^-2

Tension in the string = T =?

Acceleration in the bodies = a =?

Formula

\[T = \frac{m_{1}m_{2}}{m_{1}+m_{2}}g \]

\[a = \frac{m_{1}}{m_{1}+m_{2}}g \]

Solution

First, we will calculate the tension in the string

\[T = \frac{m_{1}m_{2}}{m_{1}+m_{2}}g \]

putting values in this equation

\[T = \frac{(24))(26))}{24 + 26}(10) \]

\[T = \frac{6540}{50} \]

T = 124.8 N

Rounding the answer

T = 125 N Ans

Now, we will calculate the acceleration in the bodies

\[a = \frac{m_{1}}{m_{1}+m_{2}}g \]

putting values in this equation

\[a = \frac{24}{24 + 26}(10)) \]

\[a = \frac{240}{50}) \]

a = 4.8 ms^-2 Ans

3.8 How much time is required to change 22 Ns momentum by a force of 20 N?

Solution

Data

Change in momentum = 𝛥P = P_f- P_i= 20 Ns

Force = F = 20 N

Time required to change the momentum = t = ?

Formula

\[F = \frac{P_{f}- P_{i}}{t} \]

Solution

\[t = \frac{P_{f}- P_{i}}{F} \]

\[t = \frac{22}{20} \]

t = 1.1 s Ans

3.9 How much is the force of friction between a wooden block of mass 5 kg and the horizontal marble floor? The coefficient of friction between wood and the marble is 0.6.

Solution

Data

Mass of block = m = 5 kg

Coefficient of friction between wood and marble = µ = 0.6

Gravitational acceleration = g = 10 ms^-1

Force of friction between wooden block and marble floor = F_s=?

Formula

F_s= µmg

Solution

F_s= µmg

Putting values in this equation

F_s= (0.6)(5)(10)

F_s= 30 N Ans

3.10 How much centripetal force is needed to make a body of mass 0.5 kg to move in a circle of radius 50 cm with a speed 3 ms^-1?

Solution

Data

Mass of body = m = 0.5 kg

Radius of Circle = r = 50 cm = 0.5 m

Speed of body = v = 3 ms^-1

Centripetal Force = F_c=?

Formula

\[F_{c} = \frac{mv^{2}}{r} \]

putting values in this equation

\[F_{c} = \frac{(0.5)(3)^{2}}{0.5} \]

\[F_{c} = \frac{(0.5)(9)}{0.5} \]

\[F_{c} = \frac{4.5}{0.5} \]

F_c = 9 N Ans

FBISE Physics MCQs for Class 9 of Unit 4 Turning Effect of Forces

4.1 Find the resultant of the following forces:

(i) 10 N along x-axis

(ii) 6 N along y-axis and

(iii) 4 N along negative x-axis.

4.2 Find the perpendicular components of a force of 50 N making an angle of 30° with x axis.

Data

Force = F = 50 N

Angle = Ѳ = 30^o

x-component of force = F_x = ?

y-component of force = F_y = ?

Formula

F_x= FcosѲ

F_y= FsinѲ

Solution

x-component of force

F_x= FcosѲ

F_x= (50)cos(30)

F_x= 43.3 N Ans

y-component of force

F_y= FsinѲ

F_y= (50)sin(30^o)

F_y= 25 N Ans

4.3 Find the magnitude and direction of a force, if its x-component is 12 N and y- component is 5 N.

Data

x-component of force = Fx = 12 N

y-component of force = Fy = 5 N

magnitude of force = F = ?

direction of force = Ѳ = ?

Formula

\[F = \sqrt{F_{x}^{2}+F_{y}^{2}}\]

\[\Theta = tan^{-1}(\frac{F_{y}}{F_{x}})\]

Solution

magnitude of force

\[F = \sqrt{F_{x}^{2}+F_{y}^{2}}\]

\[F = \sqrt{(12)^{2}+(5)^{2}}\]

\[F = \sqrt{144 + 25}\]

\[F = \sqrt{169}\]

F = 13 N Ans

direction of force

\[\Theta = tan^{-1}(\frac{F_{y}}{F_{x}})\]

\[\Theta = tan^{-1}(\frac{5}{14})\]

\[\Theta = tan^{-1}(0.416)\]

Ѳ = 22.6^owith x-axis Ans

4.4 A force of 100 N is applied perpendicularly on a spanner at a distance of 10 cm from a nut. Find the torque produced by the force.

Data

force = F = 100 N

moment arm of force = L = 0.1 m

torque produced by the force. = τ = ?

Formula

τ = FL

Solution

τ = FL

τ = (100 N)(0.1 m)

τ = 10 N Ans

4.5 A force is acting on a body making an angle of 30° with the horizontal. The horizontal component of the force is 20 N. Find the force,

Data

Angle = Ѳ = 30^o

horizontalcomponent of force = F_x = 20 N

Force = F = ?

Formula

\[F_{x}=Fcos\Theta \]

\[F = \frac{F_{x}}{Fcos\Theta }\]

Solution

\[F_{x}=Fcos\Theta \]

\[F = \frac{F_{x}}{Fcos\Theta }\]

\[F = \frac{20}{Fcos30 }\]

\[F = \frac{20}{0.866}\]

F = 23.1 N Ans

4.6 The steering of a car has a radius 16 cm. Find the torque produced by a couple of 50 N.

Data

Radius of steering of a car = r = 16 cm = 0.16 m

force = F = 50 N

torque produced by the couple. = τ = ?

Formula

The

torque of a couple = (Magnitude of one of the two forces)(perpendicular distance between them)

torque of a couple = (F)(2r)

Solution

torque of a couple = (F)(2r)

torque of a couple = (50 N)(2x0.16m)

torque of a couple = 16 Nm Ans

4.7 A picture frame is hanging by two vertical strings. The tensions in the strings are 3.8 N and 4.4 N. Find the weight of the picture frame.

Data

Tension in the first string = T₁ = 3.8 N

Tension in the second string = T₂ = 4.4 N

Weight of the picture frame = w = ?

Formula

ΣF_y = 0

Solution

ΣF_y = 0

T_{1 +}T₂- w = 0

T_{1 +}T₂ = w

w = T_{1 +}T₂

w = 3.8 N + 4.4 N

w = 8.2 N Ans

4.8 Two blocks of masses 5 kg and 3 kg are suspended by the two strings as shown. Find the tension in each string.

Data

Mass of 1^st block = m₁ = 5 kg

Mass of 2^nd block = m₂ = 3 kg

weight of 1^st block = w₁ = m₁g = (5)(10) = 50 N

weight of 2^nd block = w₂ = m₂g = (3)(10) = 30 N

Tension in the 1^st string = T₁= ?

Tension in the 2^nd string = T₂= ?

Formula

ΣF_y = 0

Solution

Tension in the 1^st string

ΣF_Y1 = 0

T₁- w₁ - w₂ = 0

T₁= w₁ + w₂

T₁= 50 N + 30 N

T₁= 80 N Ans

Tension in the 2^nd string

ΣF_Y2 = 0

T₂ - w₂ = 0

T₂= w₂

T₁= 30 N

T₂= 30 N Ans

4.9 A nut has been tightened by a force of 200 N using 10 cm long spanner. What length of a spanner is required to loosen the same nut with 150 N force?

Data

1^st force = F₁ = 200 N

Length of spanner using 1^st force = L₁ = 0.1 cm

2^nd force. = F₂ = 150 N

Length of spanner for using 2^nd force = L₂= ?

Formula

τ = FL

Solution

To loosen the same nut same torque is required from both forces i.e.

Torque produced by F₁= Torque produced by F₂

τ₁= τ₂

F₁L₁ = F₂L₂

L₂= 0.133 m Ans

4.10 A block of mass 10 kg is suspended at a distance of 20 cm from the centre of a uniform bar 1 m long. What force is required to balance it at its centre of gravity by applying the force at the other end of the bar?

First, we will draw a diagram for this problem

Data

Mass of block = m 10 kg

Weight of the block = w = mg = (10)(10) = 100 N

Distance AG = 20 cm = 0.2 m (Given in problem)

Distance GB = 50 cm = 0.5 m (From diagram)

force required to balance the rod at its centre of gravity = F =?

Formula

According to principle of moments

Clockwise moments = Anticlockwise moments

Solution

Applying principle of moments on this problem

Clockwise moments = Anticlockwise moments

w x AG = F x GB

\[F = \frac{(w)(AG)}{GB}\]

\[F = \frac{(100)(0.2)}{0.5}\]

F = 40 N Ans

FBISE Physics MCQs for Class 9 of Unit 5 Gravitation

5.1 Find the gravitational force of attraction between two spheres each of mass 1000 kg. The distance between the centres of the spheres is 0.5 m.

Data

Mass of 1^st sphere = m₁= 1000 kg

Distance between the centre of the spheres = d = 0.5 m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Gravitational force of attraction between the spheres = F = ?

Formula

\[F = G\frac{m_{1}m_{2}}{d^{2}}\]

Solution

\[F = G\frac{m_{1}m_{2}}{d^{2}}\]

Putting values in this equation

\[F = 6.67x10^{-11}\frac{(1000)(1000)}{(0.5)^{2}}\]

F= 2.67x10^-4N

5.2 The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N. Find their masses.

Data

Gravitational force between the spheres = F = 0.006673 N

Distance between the centre of the spheres = d = 0.5 m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of 1^st sphere = m₁= ?

Formula

\[F = G\frac{m_{1}m_{2}}{d^{2}}\]

Solution

\[F = G\frac{m_{1}m_{2}}{d^{2}}\]

Solve the formula for the value of m₁ m₂

_{\[m_{1}m_{2} = \frac{Fd^{2}}{G}\]}

As both the masses are identical therefore

m₁ = m_{2 =}m

Then the above formula becomes

\[m^{2} = \frac{Fd^{2}}{G}\]

Taking square root on both sides

\[m = \sqrt{\frac{Fd^{2}}{G}}\]

Putting values in this equation

\[m = \sqrt{\frac{(0.006673))(0.5))^{2}}{6.67x10^{-11}}}\]

m = 1000 kg

Because both the spheres are identical.

Mass of 1^st sphere = m₁= 1000 kg

5.3 Find the acceleration due to gravity on the surface of the Mars. The mass of Mars is 6.42x10²³ kg and its radius is 3370 km.

Data

Mass of the Mars = M_M = 6.42x10²³kg

Radius of the Mars = R_M = 3370 km = 3370x10³m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Acceleration due to gravity on the surface of the Mars = g_M = ?

Formula

\[g = \frac{GM}{R^{2}}\]

Solution

\[g_{M} = \frac{GM_{M}}{R_{M}^{2}}\]

Putting values in this equation

\[g_{M} = \frac{(6.67X10^{-11})(6.42X10^{23}))}{3370X10^{3}}\]

g = 3.77 ms^-1 Ans

5.4 The acceleration due to gravity on the surface of moon is 1.62 ms^-2. The radius of moon is 1740 km. Find the mass of moon.

Data

Acceleration due to gravity on the surface of the Moon = g_m = 1.62 ms^-2

Radius of the Moon = R_m = 1740 km = 1740x10³m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of the Mars = M_m = ?

Formula

\[g = \frac{GM}{R^{2}}\]

Solution

\[g_{m} = \frac{GM_{m}}{R_{m}^{2}}\]

rearranging this equation for the value of M_m

_{\[M_{m} = \frac{GR_{m}^{2}}{g_{m}}\]}

_{Putting values in this equation}

_{\[M_{m} = \frac{(1.62)(1.740X^{6})^{2}}{6.67X10^{-11}}\]}

M_m= 7.35x10²³ kg Ans

5.5 Calculate the value of g at a height of 3600 km above the surface of the Earth.

Data

Height above the surface of earth = h = 3600 km = 3.6x10⁶ m

Radius of the earth = R = 6400 km = 6.4x10⁶m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Acceleration due to gravity above the surface of earth = g_h = ?

Formula

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Solution

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Putting values in this equation

\[g_{h} = \frac{(6.67X10^{-11})(6X10^{24})}{(6.4X10^{6}+3.6X10^{6})^{2}}\]

g_h= 4 ms^-2 Ans

5.6 Find the value of g due to the Earth at geostationary satellite. The radius of the geostationary orbit is 48700 km.

Data

Radius of the geo stationary orbit = R = h = 48700 km = 4.8x10⁷m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Acceleration due to gravity at the geostationary satellite = g_h = ?

Formula

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Solution

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Putting values in this equation

\[g_{h} = \frac{(6.67X10^{-11})(6X10^{24})}{(4.8X^{7})^{2}}\]

g_h= 0.17 ms^-2 Ans

5.7 The value of g is 4.0 ms^-2 at a distance of 10000 km from the centre of the Earth. Find the mass of the Earth.

Data

Acceleration due to gravity above the surface of earth = g_h = 4 ms^-2

Distance from the centre of earth = R+h = 10000 km = 1x10⁷m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of the Mars = M = ?

Formula

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Solution

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

rearranging this equation for the value of M

\[M = \frac{g_{h}(R+h)^{2}}{G}\]

putting values in this equation

\[M = \frac{(4)(1X^10{7})^{2}}{6.67X10^{-11}}\]

M= 5.99x10²⁴ kg Ans

5.8 At what altitude the value of g would become one fourth than on the surface of the Earth?

Data

Acceleration due to gravity above the surface of earth = g_h = g/h = 9.8/4 = 2.45 ms^-2

Radius of earth = h = 6400 km = 6.4x10⁶m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of the earth = M = 6x10²⁴ kg

Height above the surface of earth = h = ?

Formula

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Solution

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

rearranging this equation for the value of (R+h)²

^{\[(R+h)^{2}= \frac{GM}{g_{h}}\]}

Taking square root on both sides of the above equations

\[(R+h)= \sqrt{\frac{GM}{g_{h}}}\]

rearranging this equation for the value of h

\[h= \sqrt{\frac{GM}{g_{h}}}+R\]

putting values in this equation

\[h= \sqrt{\frac{(6.67X10^{-11})(6X10^{24})}{2.45}}+6.4X10^{6}\]

h= 6.38x10⁶m Ans (This height is one earth’s radius approximately)

5.9 A polar satellite is launched at 850 km above Earth. Find its orbital
speed.

Data

Height of satellite above the surface of earth = h = 850 km = 8.5x10⁵ m

Radius of the earth = R = 6400 km = 6.4x10⁶m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Orbital speed of satellite = v_o= ?

Formula

\[v_{o}= \sqrt{g_{h}(R+h)}\]

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Solution

\[v_{o}= \sqrt{g_{h}(R+h)}\] ----------- (1)

\[g_{h} = \frac{GM}{(R+h)^{2}}\] -------------- (2)

Putting value of g_hfrom equation (2) in equation (1)

\[v_{o}= \sqrt{\frac{GM}{(R+h)^{2}}(R+h)}\]

\[v_{o}= \sqrt{\frac{GM}{(R+h)}}\]

Putting values in this equation

\[v_{o}= \sqrt{\frac{{(6.67X10^{-11})(6X10^{24})}}{6.4X10^{6}+8.5X10^{6}}}\]

v_o= 7430 ms^-1 Ans

5.10 A communication satellite is launched at 42000 km above Earth. Find its orbital speed.

Data

Height of satellite above the surface of earth = h = 42000 km = 4.2x10⁷ m

Radius of the earth = R = 6400 km = 6.4x10⁶m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Orbital speed of satellite = v_o= ?

Formula

\[v_{o}= \sqrt{g_{h}(R+h)}\]

\[g_{h} = \frac{GM}{(R+h)^{2}}\]

Solution

\[v_{o}= \sqrt{g_{h}(R+h)}\] ----------- (1)

\[g_{h} = \frac{GM}{(R+h)^{2}}\] -------------- (2)

Putting value of g_hfrom equation (2) in equation (1)

\[v_{o}= \sqrt{\frac{GM}{(R+h)^{2}}(R+h)}\]

\[v_{o}= \sqrt{\frac{GM}{(R+h)}}\]

Putting values in this equation

\[v_{o}= \sqrt{\frac{{(6.67X10^{-11})(6X10^{24})}}{6.4X10^{6}+4.2X10^{6}}}\]

v_o= 2876 ms^-1 Ans

FBISE Physics MCQs for Class 9 of Unit 6 Work and Energy

6.1 A man has pulled a cart through 35 m applying a force of 300 N. Find the work done by the man.

Data

Distance covered by the cart = S= 35 m

Force on the cart = F= 300 N

Work done by the man = W = ?

Formula

W=FS

Solution

W=FS

Putting values in this equation

W= (300 N)(35 m)

F= 10500 J Ans

6.2 A block weighing 20 N is lifted 6 m vertically upward. Calculate the potential energy stored in it.

Data

Weight of the block = w= mg = 20 N

Block is lifted to a height = h = 6 m

potential energy stored in the block = P.E = ?

Formula

P.E = mgh

Solution

P.E = mgh

Putting values in this equation

P.E = (20 N)(6 m)

P.E = 120 J Ans

6.3 A car weighing 12 kN has speed of 20 ms^-1 . Find its kinetic energy.

Data

Weight of the car = w= 12 k N = 12x10³ N = 12000 N

Speed of car = v = 20 ms^-1

Kinetic energy of the car = K.E = ?

Formula

\[K.E = \frac{1}{2}mv^{2}\]

w = mg

Solution

First, we will find the mass of the car

w = mg

\[m = \frac{w}{g}\]

Putting values in this equation

\[m = \frac{12000}{10}\]

m = 1200 kg

Now

\[K.E = \frac{1}{2}mv^{2}\]

Putting values in this equation

\[K.E = \frac{1}{2}(1200)(20)^{2}\]

K.E = 240000 J

K.E = 240 kJ Ans

6.4 A 500 g stone is thrown up with a velocity of 15ms^-1 . Find its
(i) P.E. at its maximum height
(ii) K.E. when it hits the ground

Data

mass of the stone = m= 500 g = 0.5 kg

velocity of stone = v = 15 ms^-1

gravitational constant = g = 10 ms^-1

potential energy of stone at its maximum height = P.E = ?

Kinetic energy of the stone when in hits the ground = K.E = ?

Formula

P.E = mgh\[K.E = \frac{1}{2}mv^{2}\]

\[K.E = \frac{1}{2}mv^{2}\]

Solution

Potential Energy of stone at its maximum height (P.E)

P.E = mgh ---------------- (1)

1^st we will find the maximum height “h” that stone will reach, using 3^rd equation of motion

\[2aS = v_{f}^{2} - v_{i}^{2}\]

Here

Initial velocity = v_i= 15 ms^-1

Initial velocity = v_f= 0

acceleration = a = acceleration due to gravity = g = 10

(As the stone is moving upward therefore value of g will be negative i.e., a = - g = - 10 ms^-1 )

distance = S = maximum height reached by the stone = h = ?

Putting values in 3^rd equation of motion

\[2aS = (0))^{2}- (15)^{2}\]

2(-10)h = 0 – 225

-20h = -225

h = 11.25 m

putting values in equation (1)

P.E = (0.5)(10)(11.25)

P.E = 56.25 J Ans

Kinetic Energy of stone when it hits the ground (K.E)

\[K.E = \frac{1}{2}mv^{2}\] --------------- (2)

1^st we will find the velocity with which the stone will hit the ground using 3^rd equation of motion

\[2aS = v_{f}^{2}- v_{i}^{2}\]

Here

Initial velocity = v_i= 0

Initial velocity = v_f= ?

acceleration = a = acceleration due to gravity = g = 10

distance = S = maximum height reached by the stone = h = 11.25 m

Putting values in 3^rd equation of motion

\[2gS = v_{f}^{2}- (0))^{2}\]

\[2(10)(11.25) = v_{f}^{2}- (0))^{2}\]

\[2(10)(11.25) = v_{f}^{2}\]

\[225 = v_{f}^{2}\]

\[v_{f}^{2}= \sqrt{225}\]

v_f =15 ms^-1

Putting values in this equation

\[K.E = \frac{1}{2}mv^{2}\]

\[K.E = \frac{1}{2}(0.5)(15)^{2}\]

K.E = 56.25 J Ans

Potential Energy of stone at its maximum height = P.E = 56.25 J

Kinetic Energy of the stone when in hits the ground = 56.25 J

6.5 On reaching the top of a slope 6 m high from its bottom, a cyclist has a speed of 1.5 ms^-1. Find the kinetic energy and the potential energy of the cyclist. The mass of the cyclist and his bicycle is 40 kg.

Data

Height of the slope = h = 6 m

Speed of cyclist = v = 1.5 ms^-1

mass of the cyclist and his bicycle = m= 40 kg

gravitational constant = g = 10 ms^-1

Kinetic energy of the cyclist = K.E = ?

potential energy of cyclist = P.E = ?

Formula

\[K.E = \frac{1}{2}mv^{2}\]

P.E = mgh

Solution

Kinetic Energy of stone when it hits the ground (K.E)

\[K.E = \frac{1}{2}mv^{2}\]

Putting values in this equation

\[K.E = \frac{1}{2}(40)(1.5)^{2}\]

\[K.E = \frac{1}{2}(40)(2.25)\]

K.E = 45 J Ans

Potential Energy of stone at its maximum height (P.E)

P.E = mgh

Putting values in this equation

P.E = (40)(10)(6)

P.E = mgh

P.E = 2400 J Ans

6.6 A motorboat moves at a steady speed of 4 ms^-1 . Water resistance acting on it is 4000 N. Calculate the power of its engine.

Data

Speed of motorboat = v = 4 ms^-1

Water resistance acting on the boat = F= 4000 N

Power of the engine of boat = P = ?

Formula

\[P = \frac{W}{t}\]

W = FS

Solution

\[P = \frac{W}{t}\]

W =FS

Putting this value in above equation

\[P = \frac{FS}{t}\]

\[P = F(\frac{S}{t})\

P = Fv

Putting values in this equation

P = (4000)(4)

P = 16000 W

P = 16x10³ W

P = 16 kW Ans

6.7 A man pulls a block with a force of 300 N through 50 m in 60 s. Find the power used by him to pull the block.

Data

Force applied by the man = F= 300 N

Distance covered by the block = 50 m

Time of application of force = t = 60 s

Power applied by the man = P = ?

Formula

\[P = \frac{W}{t}\]

W = FS

Solution

\[P = \frac{W}{t}\]

W =FS

Putting this value in above equation

\[P = \frac{FS}{t}\]

Putting values in this equation

\[P = \frac{FS}{t}\]

\[P = \frac{(300)(50)}{60}\]

P = 250 W Ans

6.8 A 50 kg man moved 25 steps up in 20 seconds. Find his power if each step is 16 cm
high.

Data

Mass of the man = m = 50 kg

Number of steps = N = 25

Height of each step = 𝛥h = 16 cm = 0.16 m

height=h=(Number of steps)(height of each step)

height=h=(N)( 𝛥h)

height=h=(25)( 0.16)

height=h= 4 m

Time = t = 20 s

Power of the man = P = ?

Formula

\[P = \frac{W}{t}\]

W = FS

Solution

\[P = \frac{W}{t}\]

W =FS = wh = mgh

Putting this value in above equation

\[P = \frac{mgh}{t}\]

Putting values in this equation

\[P = \frac{(50)(10)(4)}{20}\]

P = 100 W Ans

6.9 Calculate the power of a pump which can lift 200 kg of water through a height of
6 m in 10 seconds.

Data

Mass of water = m = 200 kg

height=h= 6 m

Time = t = 10 s

Power of the man = P = ?

Formula

\[P = \frac{W}{t}\]

W = FS

Solution

\[P = \frac{W}{t}\]

W =FS = wh = mgh

Putting this value in above equation

\[P = \frac{mgh}{t}\]

Putting values in this equation

\[P = \frac{(200)(10)(6)}{10}\]

P = 1200 W Ans

6.10 An electric motor of 1hp is used to run water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 litres and height of 15 m. Find the actual work done by the electric motor to fill the tank. Also find the efficiency of the system.
(Density of water = 1000 kgm^-3)
(Mass of 1 litre of water = 1 kg)

Data

Power of electric motor = 1hp = 746 W

Time taken by the pump to fill the tank = t = 10 minutes = (10)(60s) = 600 s

Capacity of tank to store water in kilograms = m = 800 kg

Height of the tank = 15 m

Actual work done by the motor to fill the tank = W = ?

Efficiency of the system = ?

Formula

\[P = \frac{W}{t}\]

\[%efficiency = \frac{required form of output}{total input energy}X100\]

Solution

Actual work done by the motor to fill the tank

\[P = \frac{W}{t}\]

W = P x t

Putting values in this equation

W = 746 x 600

W = 447600 J Ans

Efficiency of the system

\[%efficiency = \frac{required form of output}{total input energy}X100\]

Here

Input energy = actual work done by the motor = W = 447600 J

Output work = mgh = (800)(10)(15) = 120000 J

Putting values in the above equation

\[%efficiency = \frac{447600}{120000}X100\]

% efficiency = 26.8 % Ans

FBISE Physics MCQs for Class 9 of Unit 7 Properties of Matter

7.1 A wooden block measuring40 cm x 10 cm x 5 cm has a mass 850 g. Find the density of wood.

Data

Volume of the wooden block = V = 40 cm x 10 cm x 5 cm

= 0.40 m x 0.10 m x 0.05 m = 2x10^-3 m³.

Mass of wooden block= m= 850 g = 0.850 kg

Density of the wood = ϱ = ?

Formula

Solution

Putting values in this equation

ϱ = 425 kgm^-3

7.2 How much would be the volume of ice formed by freezing 1 litre of water?

Data

Volume of water = V₁= 1 litre

Volume of ice = V₂= ?

Density of water = ϱ₁= 1000 kgm^-3

Density of water = ϱ₂= 920 kgm^-3

Formula

Solution

V₂= 1.09 X 1

V₂= 1.09 litre Ans

7.3 Calculate the volume of the following objects:

(i) An iron sphere of mass 5 kg, the density of iron is 8200 kgm^-3.

(ii) 200 g of lead shot having density 11300 kgm^-3.

(iii) A gold bar of mass 0.2 kg. The density of gold is 19300 kgm^-3.

(i) An iron sphere of mass 5 kg, the density of iron is 8200 kgm^-3.

Data

mass of Iron sphere = m= 5 kg

density of iron = ϱ = 8200 kgm^-3

volume of iron sphere = V = ?

Formula

Solution

Rearranging this equation

Putting values in this equation

V= 6.1X10^-4 m³ Ans

(ii) 200 g of lead shot having density 11300 kgm^-3

Data

mass of lead shot = m= 200g = 0.2 kg

density of lead = ϱ = 11300 kgm^-3

volume of lead shot = V = ?

Formula

Solution

Rearranging this equation

Putting values in this equation

V= 1.77X10^-5 m³ Ans

(iii) A gold bar of mass 0.2 kg. The density of gold is 19300 kgm^-3

Data

mass of gold bar = m= 0.2 kg

density of iron = ϱ = 19300 kgm^-3

volume of iron sphere = V = ?

Formula

Solution

Rearranging this equation

Putting values in this equation

V= 1.04X10^-5 m³ Ans

7.4 The density of air is 1.3 kgm^-3. Find the mass of air in a room measuring 8m x 5m x 4m.

Data

density of air = ϱ = 1.3 kgm^-3

volume of room = V = 8m x 5m x 4m = 160 m³

mass of air in the room = m= ?

Formula

Solution

Rearranging this equation

m = ϱ X V

Putting values in this equation

m = (1.3)(160)

m= 208 kg Ans

7.5 A student presses her palm by her thumb with a force of 75 N. How much would be the pressure under her thumb having contact area 1.5 cm² ?

Data

Force applied by the student = 75 N

Contact area under the thumb = A = 1.5 cm²= 1.5x10^-4m²

Pressure under the thumb = P = ?

Formula

Solution

Putting values in this equation

P = 5x10⁵ Nm^-2 Ans

7.6 The head of a pin is a square of side 10 mm. Find the pressure on it due to a force of 20 N.

Data

Length of one side of head of pin = L= 10 mm = 10x10^-3 m

Area of head of the pin = A = (L)(L) = L²= (10x10^-3 m)² = 1x10^-4

Force acting on the pin = F = 20 N

Pressure on the pin = P = ?

Formula

Solution

Putting values in this equation

P = 2x10⁵ Nm^-2 Ans

7.7 A uniform rectangular block of wood 20 cm x 7.5 cm x 7.5 cm and of mass 1000g stands on a horizontal surface with its longest edge vertical. Find

(i) the pressure exerted by the block on the surface

(ii) density of the wood.

Data

Volume of wooden block = V= 20 cm x 7.5 cm x 7.5 cm = 0.20 m x 0.075 m x 0.075 cm

= 1.125x10^-3 m³

Area of block in contact with surface = A = 7.5 cm x 7.5 cm = (0.075m)( 0.075m)

= 5.625x10^-3 m²

Mass of wooden block = m= 1000 g = 1 kg

the pressure exerted by the block on the surface = P = ?

density of the wood= ϱ =?

(i) the pressure exerted by the block on the surface

Formula

F = w = mg

Solution

No horizontal surface is equal to the w we know that the force exerted by the block on the weight of the block i.e.

F = w = mg

Putting F from this equation in the above equation

P = 1778 Nm^-2 Ans

(ii) density of the wood.

Formula

Solution

Putting values in this equation

ϱ = 889 kgm^-3

7.8 A cube of glass of 5 cm side and mass 306 g has a cavity inside it. If the density of glass is 2.55 gcm^-3. Find the volume of the cavity.

Data

Length of one side of glass cube = L = 5 cm = 5x10^-2 m

Mass of glass cube with cavity = m = 306 g= 0.306 kg

Density of glass = ϱ = 2.55 gcm^-3 = 2550 kgm^-3

Volume of cavity = V_Cavity = ?

Total volume of glass cube = V_Cube = (L)(L)(L) = L³ = (5x10^-2 m)³ = 2.5x10^-3m³

Formula

Volume of Cube = Length x width x height

Solution

Volume of cube without cavity = V_Cube(Without Cavity) = (L)(L)(L) = L³

= (5x10^-2 m)³

= 1.25x10^-4m³

Volume of cube with cavity =

V_Cube(With Cavity) = 1.2x10^-4m³

Volume of Cavity = Volume of cube without cavity - Volume of cube with cavity

V_Cavity = V_Cube(Without Cavity) - V_Cube(With Cavity)

V_Cavity = 1.25x10^-4m³ - 1.2x10^-4m³

V_Cavity = 5 x 10^-6 m³

V_Cavity = 5 cm³

7.9 An object has weight 18 N in air. Its weight is found to be 11.4 N when immersed in water. Calculate its density. Can you guess the material of the object?

Data

Weight of object in air = w₁= 18 N

Weight of object in water = w₂= 11.4 N

Density of liquid = ϱ = 1000 kgm^-3

Density of the object = D = ?

Material of object = ?

Formula

Solution

Putting values in this equation

D = 2727 kgm^-3 Ans

7.10 A solid block of wood of density 0.6 gcm^-3 weighs 3.06 N in air. Determine

a) volume of the block

b) the volume of the block immersed when placed freely in a liquid of density 0.9 gcm^-3 ?

Density of wooden block = ϱ₁= 0.6 gcm^-3 = 600 kgm^-3

Weight of the block = w = 3.06 N

Mass of the block = m = w/g = 3.06/10 = 0.306 kg

Density of wooden block = ϱ₂= 0.9 gcm^-3 = 900 kgm^-3

Volume of block immersed in liquid = V₂= ?

a) volume of the block

Formula

Solution

Rearranging this equation

Putting values in this equation

V₁= 5.1X10^-4 m³

V₁= 5.1X10^-4 X 10⁶ cm³

V₁= 510 cm³ Ans

b) the volume of the block immersed when placed freely in a liquid of density 0.9 gcm^-3 ?

Formula

Upthrust = ϱgV

Solution

Upthrust = ϱ₂gV₂

w = ϱ₂gV₂

Rearranging this equation for value of V₂

Putting values in this equation

V₁= 3.4X10^-4 m³

V₁= 3.4X10^-4 X 10⁶ cm³

V₁= 340 cm³ Ans

7.11 The diameter of the piston of a hydraulic press is 30 cm. How much force is required to lift a car weighing 20 000 N on its piston if the diameter of the piston of the pump is 3 cm?

Data

Diameter of piston of hydraulic press = D= 30 cm = 0.3 m

Weight of the car = F₁= 20000 N

Diameter of piston of the pump = d= 3 cm = 0.03 m

Force required to lift the car = F₂ = ?

Formula

Solution

First, we calculate the areas of the pistons

Area of piston of hydraulic press

Putting values in this equation

A= 0.07065 m²

Area of piston of pump

Putting values in this equation

a= 7.065X10^-4 m²

Now using formula of Hydraulic press

Rearranging this equation for F₁

Putting values in this equation

F₁= 200 N

7.12 A steel wire of cross-sectional area 2x10^-5 m² is stretched through 2 mm by a force of 4000 N. Find the Young's modulus of the wire. The length of the wire is 2 m.

Data

Cross-sectional Area of steel wire = A= 2x10^-5 m²

Length of wire through which it stretched = 𝛥L = 2 mm = 2x10^-3m

Force applied on the wire = F = 4000 N

Original length length of the wire = L_o = 2 m

Young's modulus of the wire = Y = ?

Formula

Solution

Putting values in this equation

Y = 2x10¹¹ Nm^-2 Ans

FBISE Physics MCQs for Class 9 of Unit 8 Thermal Properties of Matter

8.1 Temperature of water in a beaker is 50°C. What is its value in Fahrenheit scale?

Data

Temperature of water in Centigrade = T_c = 50 °C

Temperature of water in Fahrenheit = T_F = ?

Formula

T_F = (1.8)T_c + 32

Solution

T_F = (1.8)T_c + 32

Putting values in this equation

T_F = (1.8)(50 °C) + 32

T_F = 122 °F Ans

8.2 Normal human body temperature is 98.6°F. Convert it into Celsius scale and Kelvin scale

Temperature of Human body in Fahrenheit = T_F = 98.6 °C

Temperature of Human Body in Celsius = T_C = ?

Temperature of Human Body in Kelvin = T_K = ?

Formula

T_K = T_c + 273

Solution

Putting values in this equation

T_c= 37 °C Ans

T_K = T_c + 273

Putting values in this equation

T_F = 371.6 K Ans

8.3 Calculate the increase in the length of an aluminium bar 2 m long when heated from 0°C to 20°C. If the thermal coefficient of linear expansion of aluminium is 2.5x10^-5K^-1

Data

Length of Aluminium bar = L_o= 2 m

Initial Temperature of the bar = T_o = 0 °C = 0 °C + 273 = 273 K

Final temperature of the bar = T = 20 °C = 20 °C + 273 = 293 K

Change in temperature of the block = 𝛥T = T - T_o = 293 K – 273 K = 20 K

Thermal coefficient of linear expansion of aluminium = α = 2.5x10^-5K^-1

Increase in length of aluminium = 𝛥L = ?

Formula

𝛥L = α L_o𝛥T

Solution

𝛥L = α L_o𝛥T

Putting values in this equation

𝛥L = (2.5x10^-5K^-1)(2m)(20)

𝛥L = 0.001 m

𝛥L = 0.1 cm Ans

8.4 A balloon contains 1.2 m³ air at 15 °C. Find its volume at 40 °C. Thermal coefficient of volume expansion of air is 3.67x10^-3 K^-1.

Data

Volume of air in balloon = V_o= 1.2 m³

Initial Temperature of the bar = T_o = 15 °C = 15 °C + 273 = 288 K

Final temperature of the bar = T = 40 °C = 40 °C + 273 = 313 K

Change in temperature of the block = 𝛥T = T - T_o = 313 K – 288 K = 25 K

Thermal coefficient of volume expansion of air = β = 3.67x10^-3 K^-1

Volume of balloon at 40 °C = V = ?

Formula

V = V_o(1 + β𝛥T)

Solution

V = V_o(1 + β𝛥T)

Putting values in this equation

V = (1.2){1+ (3.67x10^-3)(25)}

V= 1.3 m³ Ans

8.5 How much heat is required to increase the temperature of 0.5 kg of water from 10 °C to 65 °C?

Data

Mass of water = 0.5 kg

Initial Temperature of the bar = T_o = 10 °C = 10 °C + 273 = 283 K

Final temperature of the bar = T = 65 °C = 65 °C + 273 = 338 K

Change in temperature of the block = 𝛥T = T - T_o = 338 K – 283 K = 55 K

Specific heat of water = c = 4200 Jkg^-1K^-1

Amount of Heat required = Q = ?

Formula

Q = cm 𝛥T

Solution

Q = cm 𝛥T

Putting values in this equation

Q = (4200)(0.5)(55)

Q = 115500 J Ans

8.6 An electric heater supplies heat at the rate of 1000 joule per second. How much time is required to raise the temperature of 200 g of water from 20 °C to 90 °C?

Data

Rate of heat supplied = P= Q/t = 1000 Js^-1

Mass of water = m = 200 g = 0.2 kg

Initial Temperature of the bar = T_o = 20 °C = 20 °C + 273 = 293 K

Final temperature of the bar = T = 90 °C = 90 °C + 273 = 363 K

Change in temperature of the block = 𝛥T = T - T_o = 363 K – 293 K = 70 K

Specific heat of water = c = 4200 Jkg^-1K^-1

Amount of Heat required = Q = ?

Time required to raise the temperature = t = ?

Formula

Q = cm 𝛥T

P = Q/t

Solution

Q = cm 𝛥T

Putting values in this equation

Q = (4200)(0.2)(70)

Q = 58800 J

Now

P = Q/t

P x t = Q

t = Q/P

t = 58800/1000

t = 58.8 s

8.7 How much ice will melt by 50000 J of heat? Latent heat of fusion of ice = 336000 J kg^-1

Data

Amount of heat = Q = 50000 J

Latent heat of fusion of ice = H_f = 336000 Jkg^-1

Mass of ice = m = ?

Formula

Q = m H_f

Solution

Q = m H_f

m = 0.15 kg

m = 150 g Ans

8.8 Find the quantity of heat needed to melt 100g of ice at -10 °C into water at 10 °C. (39900 J)

(Note: Specific heat of ice is 2100 Jkg^-1K^-1, specific heat of water is 4200 Jkg^-1K^-1.Latent heat of fusion of ice is 336000 Jkg^-1).

Data

Mass of ice = m = 100 g = 0.1 kg

Initial Temperature of ice = T_o = -10 °C = -10 °C + 273 = 263 K

Final temperature of water = T = 10 °C = 00 °C + 273 = 283 K

Specific heat of water = c = 2100 Jkg^-1K^-1

Specific heat of water = c = 4200 Jkg^-1K^-1

Latent heat of fusion of ice = H_f = 336000 Jkg^-1

Amount of Heat required = Q = ?

Formula

Q = cm 𝛥T

Q = m H_f

Solution

Q = Q₁+ Q₂ + Q₃

Here

Q = Total Heat

Q₁ = Heat required to raise the temperature of ice from -10 °C to 0 °C

Q₂= Heat required to melt the ice at 0 °C

Q₃= Heat required to raise the temperature of ice from 0 °C to 10 °C

Now

Q_{1 =}cm 𝛥T

Putting values in this equation

Q_{1 =}(2100)(0.1)(10)

Q_{1 =}2100 J

Now

Q₂= mH_f

Q₂= (0.1)(336000)

Q₂= 33600 J

Now

Q_{3 =}cm 𝛥T

Putting values in this equation

Q_{3 =}(4200)(0.1)(10)

Q_{1 =}4200 J

Putting these values in the above equation

Q = 2100 + 33600 + 4200

Q = 39900 J

8.9 How much heat is required to change 100 g of water at 100°C into steam?

(Latent heat of vaporization of water is 2.26x10⁶ Jkg^-1).

Data

Mass of ice = m = 100 g = 0.1 kg

Latent heat of vaporization of water = H_v = 2.26x10⁶ Jkg^-1

Amount of heat = Q_v = ?

Formula

Q_v = m H_v

Solution

Q_v = m H_v

Q_v = (0.1)( 2.26x10⁶ )

Q_v = 2.26x10⁵ J

8.10 Find the temperature of water after passing 5 g of steam at 100 °C through 500 g of water at10 °C.

(Note: Specific heat of water is 4200 Jkg^-1K^-1, Latent heat of vaporization of water is 2.26 x10⁶ Jkg^-1).

Data

Mass of steam = m₁ = 5 g = 0.005 kg

Temperature of steam = T₁ = 100 °C = 100 °C + 273 = 373 K

Mass of water = m₂ = 500 g = 0.5 kg

Temperature of water = T₂ = 10 °C = 10 °C + 273 = 283 K

Specific heat of water = c = 4200 Jkg^-1K^-1

Latent heat of vaporization of water = H_v = 2.26 x10⁶ Jkg^-1

Final mass of water whose temperature to be measured = m = m_1
+m₂

= 0.005kg + 0.5kg

= 0.505 kg

Final Temperature of water = T = ?

Formula

Q = cm 𝛥T

Q = m H_f

Solution

Heat Loss by the steam = Heat gain by the water

m₁H_v + cm₁𝛥T = cm𝛥T

(0.005)( 2.26 x10⁶) + (4200)(0.005)(T - 373) = (4200)(0.505)(T - 283)

11300 + 21(T - 373) = 2121(T - 283)

11300 + 21T – 7833 = 2121T – 600243

600243 + 11300 – 7833 = 2121T – 21T

603710 = 2100T

T = 603710/2100

T = 287.48 K

T = 287.48 – 273 °C

11300 - 21T + 7833 = 2121T – 600243

600243 + 11300 + 7833 = 2121T + 21T

619376 = 2142T

T = 619376/2142

T = 289.15 K

T = 289.15 – 273 °C

T = 16.15 °C

T = 16.2 °C Ans

Heat Loss by the steam = m₁H_f + cm₁𝛥T

Heat Loss by the steam = (0.005)( 2.26 x10⁶) + (4200)(0.005)(T - 373)

Heat gain by the water = cm₂𝛥T

Heat gain by the water = (4200)(0.5)(T - 283)

Here

Q = Total Heat

Q₁ = Heat required to raise the temperature of ice from -10 °C to 0 °C

Q₂= Heat required to melt the ice at 0 °C

Q₃= Heat required to raise the temperature of ice from 0 °C to 10 °C

Now

Q_{1 =}cm 𝛥T

Putting values in this equation

Q_{1 =}(2100)(0.1)(10)

Q_{1 =}2100 J

Now

Q₂= mH_f

Q₂= (0.1)(336000)

Q₂= 33600 J

Now

Q_{3 =}cm 𝛥T

Putting values in this equation

Q_{3 =}(4200)(0.1)(10)

Q_{1 =}4200 J

Putting these values in the above equation

Q = 2100 + 33600 + 4200

Q = 39900 J

FBISE Physics MCQs for Class 9 of Unit 9 Transfer of Heat

9.1 The concrete roof of a house of thickness 20 cm has an area 200 m². The temperature inside the house is 15 °C and outside is 35°C. Find the rate at which thermal energy will be conducted through the roof. The value of k for concrete is 0.65 Wm^-1K^-1.

Data

Thickness of wall = L = 20 cm = 0.20 m

Area of wall = A = 200 m²

Temperature outside of house = T₁ = 35 °C = 35 + 273 K = 308 K

Temperature inside the house = T₂ = 15 °C = 15 + 273 K = 288 K

Thermal conductivity of wall = k = 0.65 Wm^-1K^-1

Rate at which thermal energy conducted through the roof = Q/t = ?

Formula

Solution

Putting values in this equation

Q/t = 13000 Js^-1 Ans

9.2 How much heat is lost in an hour through a glass window measuring 2.0 m by 2.5 m when inside temperature is 25 °C and that of outside is 5°C, the thickness of glass is 0.8 cm and the value of k for glass is 0.8 Wm^-1K^-1.

Data

Thickness of glass = L = 0.8 cm = 0.008 m

Area of wall = A = 2m x 2.5m = 5 m²

Temperature outside of house = T₁ = 25 °C = 35 + 273 K = 298 K

Temperature inside the house = T₂ = 05 °C = 05 + 273 K = 278 K

Time = t = 1 h = 1 x 60 x 60 = 3600 s

Thermal conductivity of wall = k = 0.8 Wm^-1K^-1

Heat lost in an through the glass window = Q = ?

Formula

Solution

Putting values in this equation

Q = 3.6x10⁷ J Ans

FBISE Physics MCQs for Class 10

Prepare for your FBISE Physics exam of Class 10 with our MCQs practice and tips. Learn the best strategies for answering multiple choice questions and boost your scores!

FBISE Physics MCQs for Class 10 of Unit 10 Simple Harmonic Motion and Waves

FBISE Physics MCQs for Class 10 of Unit 11 Sound

FBISE Physics MCQs for Class 10 of Unit 12 Geometrical Optics

FBISE Physics MCQs for Class 10 of Unit 13: Electrostatics

FBISE Physic MCQs for Class 10 of Unit 14 Current Electricity

FBISE Physics MCQs for Class 10 of Unit 15 Electromagnetism

FBISE Physics MCQs for Class 10 of Unit 16 Basic Electronics

FBISE Physics MCQs for Class 10 of Unit 10 Information and Communication Technology

FBISE Physics MCQs for Class 10 of Unit 18 Atomic and Nuclear Physics

FBISE Physics MCQs for Class 11

Prepare for your FBISE Physics exam of Class 11 with our MCQs practice and tips. Learn the best strategies for answering multiple choice questions and boost your scores!

FBISE Physics MCQs for Class 11 of Unit 1 Measurements

FBISE Physics MCQs for Class 11 of Unit 2 Vectors and Equilibrium

FBISE Physics MCQs for Class 11 of Unit 3 Motion and Force

FBISE Physics MCQs for Class 11 of Unit 4 Work and Energy

FBISE Physics MCQs for Class 11 of Unit 5 Circular Motion

FBISE Physics MCQs for Class 11 of Unit 6 Fluid Dynamics

FBISE Physics MCQs for Class 11 of Unit 7 Oscillations

FBISE Physics MCQs for Class 11 of Unit 8 Waves

FBISE Physics MCQs for Class 11 of Unit 9 Physical Optics

FBISE Physics MCQs for Class 11 of Unit 10 Optical Instruments

FBISE Physics MCQs for Class 11 of Unit 11 Heat and Thermodynamics

FBISE Physics MCQs for Class 12

Prepare for your FBISE Physics exam of Class 12 with our MCQs practice and tips. Learn the best strategies for answering multiple choice questions and boost your scores!