Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. the work done by the force of gravity during the time the particle goes up is

  • 0.5 J

  • -0.5 J

  • −1.25 J

  • 1.25 J


C.

−1.25 J

negative mgh space equals negative space mg open parentheses fraction numerator straight v squared over denominator 2 straight g end fraction close parentheses space equals space minus space 1.25 space straight J
270 Views

2.

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms−2:

  • 2.45 ×10−3 kg

  • 6.45 x×10−3 kg

  • 9.89 ×10−3 kg

  • 12.89 ×10−3 kg


D.

12.89 ×10−3 kg

Given potential energy burnt by lifting weight

= mgh = 10 x 9.8 x 1 x 1000 = 9.8 x 104 J

If mass lost by a person be m, then energy dissipated

 = m x 2 x 38 x 107 J /10

⇒ m = 5 x 10-3 x 9.8 / 3.8

= 12.89 x 10-3 kg 

628 Views

3.

An athlete in the Olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

  • 200 J − 500 J

  • 2 × 105J − 3 × 105J

  •  20,000 J − 50,000 J

  • 2,000 J − 5,000 J


C.

 20,000 J − 50,000 J

Approximate mass = 60 kg Approximate
velocity = 10 m/s Approximate
KE = (1/2)x60 x1000 = 3000 J
= KE range ⇒ 2000 to 5000 joule

346 Views

4.

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly. (Surface tension of soap solution = 0.03 Nm-1)

  • 4π mJ 

  • 0.2π mJ

  • 2π mJ

  • 0.4π mJ


D.

0.4π mJ

Work done = Change in surface energy
⇒ W = 2T x 4π (R22-R12)
 = 2 x 0.03 x 4π [ (5)2-(3)2] x 10-4
 = 0.4 π mJ

627 Views

5.

The potential energy of a 1 kg particle free move along the x-axis is given by
straight V left parenthesis straight x right parenthesis space equals space open parentheses straight x to the power of 4 over 4 minus straight x squared over 2 close parentheses space straight J

The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is

  • 2

  • 3 divided by square root of 2
  • square root of 2
  • 1 divided by square root of 2

B.

3 divided by square root of 2
kE subscript max space equals space straight E subscript straight T minus straight U subscript min
straight U subscript min space left parenthesis plus-or-minus 1 right parenthesis space equals negative 1 fourth straight J
KE subscript max space equals space 9 divided by 4 space straight J thin space
rightwards double arrow space straight U space equals space fraction numerator 3 over denominator square root of 2 end fraction space straight J
349 Views

6.

This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.
If two springs S1 and S2 of force constants k1 and k2, respectively, are stretched by the same force, it is found that more work is done on spring S1 than on spring S2.

Statement 1: If stretched by the same amount, work done on S1, will be more than that on S2
Statement 2 : k1 < k2

  • Statement 1 is false, Statement 2 is true

  • Statement 1 is true, Statement 2 is false

  • Statement 1 is true, Statement 2 is the correct explanation for statement 1

  • Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for statement 1.


A.

Statement 1 is false, Statement 2 is true

As no relation between k1 and k2 is given in the question, that is why nothing can be predicted about the statement I. But as in Statment II. k1<k2
Then, for same force

straight W space equals space straight F. straight x space equals space straight F. straight F over straight k space equals space straight F squared over straight k
straight W space proportional to space 1 over straight k space straight i. straight e space for space constant space straight F
straight W subscript 1 over straight W subscript 2 space equals straight K subscript 2 over straight K subscript 1
But space for space same space displacement
straight W space equals space straight F. straight x space equals 1 half space kx. straight x space equals space 1 half kx squared
straight W space proportional to space straight K comma space straight W subscript 1 over straight W subscript 2 space equals space straight K subscript 1 over straight K subscript 2
straight W subscript 1 space less than straight W subscript 2

698 Views

7.

A force straight F with rightwards arrow on top space equals space left parenthesis 5 straight i with hat on top space plus 3 straight j with hat on top space plus 2 straight k with hat on top right parenthesis space straight N is applied over a particle which displaces it from its origin to the straight r with rightwards arrow on top space equals space left parenthesis 2 straight i with hat on top minus straight j with hat on top right parenthesis space straight m.The work done on the particle in joules is

  • -7

  • +7

  • +10

  • +13


B.

+7

Work done in displacing the particle
straight W space equals space straight F with rightwards arrow on top space. straight r with rightwards arrow on top
space equals space left parenthesis 5 straight i with hat on top space plus space 3 straight j with hat on top space plus 2 straight k with hat on top right parenthesis. left parenthesis 2 straight i with hat on top space minus straight j with hat on top right parenthesis
space equals space 5 space straight x space 2 space plus 3 space straight x space left parenthesis negative 1 right parenthesis space plus space 2 straight x 0
space equals space 10 minus 3
equals space 7 straight J

209 Views

8.

A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

  • 7.2 J

  • 3.6 J

  • 120 J

  • 1200 J 


B.

3.6 J

Mass per length
= M/L
= 4/2 = 2 kg/m
The mass of 0.6 m of chain = 0.6 x 2 = 1.2 kg
The centre of mass of hanging part = 0.6 +0 /2 = 0.3 m
Hence, work done in pulling the chain on the table
W =mgh
= 1.2 x 10 x 0.3
= 3.6 J

266 Views

9.

A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

  • 40 m/s

  • 20 m/s

  • 10 m/s

  • 10 square root of 30 m/s


A.

40 m/s

mgh space equals space 1 half mv squared
straight v equals space square root of 2 gh end root
space equals square root of 2 space straight x space 10 space straight x 80 end root space equals 40 space straight m divided by straight s
349 Views

10.

When a rubber band is strecthed by a distance x, it exerts a restoring force of magnitude F = ax +bx2, where a and b are constants. The work done in stretching are unstretched rubber-band by L is

  • aL2 +bL2

  • 1 half left parenthesis aL squared plus bL cubed right parenthesis
  • aL squared over 2 space plus bL cubed over 3
  • 1 half space open parentheses aL squared over 2 plus bL cubed over 3 close parentheses

C.

aL squared over 2 space plus bL cubed over 3
straight U subscript straight f minus straight U subscript straight i space equals space minus straight W space equals space minus space integral subscript straight i superscript straight f straight F. dr
Given, F = ax +bx2
We know that work done in stretching the rubber band by L is 
|dW|= |Fdx|
vertical line straight W vertical line space equals space integral subscript 0 superscript straight L left parenthesis ax space plus bx squared right parenthesis dx
space equals space open square brackets ax squared over 2 close square brackets subscript straight O superscript straight L space plus space open square brackets bx cubed over 3 close square brackets subscript straight O superscript straight L
equals open square brackets aL squared over 2 minus fraction numerator ax space left parenthesis 0 right parenthesis squared over denominator 2 end fraction close square brackets space plus space open square brackets fraction numerator straight b space straight x space straight L cubed over denominator 3 end fraction minus fraction numerator straight b space straight x space left parenthesis 0 right parenthesis cubed over denominator 3 end fraction close square brackets
vertical line straight W vertical line space equals space aL squared over 2 plus bL cubed over 3
368 Views