This set of Physics Aptitude Test for NEET Exam focuses on “Work Done by a Variable Force”.

1. A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in the figure. What is the work done?

a) 20 J

b) 40 J

c) 60 J

d) 80 J

View Answer

Explanation: The work done (W) is given by;

W = Force x Displacement

The area enclosed by the curves in the graph shown gives the work done;

Enclosed area = [(1/2) x 4 x 10] + (4 x 10) + [(1/2) x 4 x 10]

= 80 Nm

= 80 J.

2. The length of a smooth inclined plane of 30-degree inclination is 5 m. The work done in moving a 10 kg mass from the bottom of the inclined plane to the top is _____ Joules. (Assume g = 10m/s^{2})

a) 250

b) 1000

c) 1250

d) 500

View Answer

Explanation: Length (l) = 5 m

Height = Displacement = 5 x sin30

= 5 x (1/2)

= 2.5

Force on object = m x g

= 10 X 10

= 100 N

Work = Force x Displacement

= 100 x 2.5

= 250 Joules.

3. A rope has a uniform mass density of 0.4 kg/m. The rope is 10 m long and is lowered into a 10 m deep pit. The bottom part of the rope just touches the bottom of the pit. What is the work done to pull the rope out of the well completely? (Assume g = 10m/s^{2})

a) 200 J

b) 400 J

c) 2000 J

d) 4000 J

View Answer

Explanation: Mass density (p) = 0.4 kg/m

Length (l) = 10 m

Force = f(x)

= m(x)*g

= 0.4x*g

= 4x

Displacement (d) = 10 m

Work = \(\int_{0}^{l}\)f(x)dx

= \(\int_{0}^{10}\)4xdx

= 200 J.

4. A bucket filled with water weighing 20 kg is raised from a well of depth 20 m. If the linear density of the rope is 0.2 kg/m, the work done is _____ (Assume g = 10m/s^{2})

a) 4000 J

b) 4040 J

c) 4400 J

d) 4800 J

View Answer

Explanation: Mass density (p) = 0.2 kg/m

Length (l) = 20 m

Force = f(x)

= m(x)*g

= 0.2x*g

= 2x

Displacement (d) = 20 m

Work = \(\int_{0}^{l}\)f(x)dx+ (20*g*20)

= \(\int_{0}^{20}\)2xdx + 4000

= 4400 J.

5. A force F = (5i – 3j +2k) N moves a particle from r1 = (2i + 7j + 4k) m to r2 = (5i + 2j + 8k) m. What is the work done by the force?

a) 18 J

b) 28 J

c) 38 J

d) 48 J

View Answer

Explanation: Displacement (d) = r2 – r1

= (3i – 5j + 4k) m

Work = Force . Displacement; [Dot product of vectors]

= (5i – 3j +2k) .(3i – 5j + 4k)

= 38 J.

6. A person of mass 50 kg carrying a load of 20 kg walks up a staircase. The width and height of each step are 0.25 m and 0.2 m respectively. What is the work done by the man in walking 20 steps?(Assume g = 10m/s^{2})

a) 800 J

b) 1600 J

c) 2000 J

d) 2800 J

View Answer

Explanation: Total mass (m) = 50 + 20 = 70 kg

Vertical displacement = 0.2 x 20 = 4m

Horizontal displacement = 0.25 x 20 = 5m

Displacement = (5i + 4j) m

Force = m x acceleration

= 70 x (0i + 10j)

= (700j) N

Work = Force .Displacement

= (0i + 700j) . (5i + 4j)

= 2800 J.

7. A chain of length 4 m is kept on a table. 2 m of the chain hangs freely from the table’s edge. The mass of the whole chain is 8 kg. What is the work done to pull the whole chain onto the table? (Assume g = 10m/s^{2})

a) 20 J

b) 40 J

c) 60 J

d) 80 J

View Answer

Explanation: Mass per length of chain (p) = 8/4 = 2 kg/m

Mass of ‘x’ m of length = 2x kg

Force = \(\int_{0}^{2}\)(2x * g)dx

= \(\int_{0}^{2}\)20xdx

= 40 J.

8. The area under the force (F) versus displacement (x) graph gives work.

a) True

b) False

View Answer

Explanation: The area under this graph would be \(\int_{x1}^{x2}\)F * dx, which is work. Here x1 & x2 are two positions on the x-axis. Hence the statement is true.

9. An object of mass 1kg is subjected to a variable force in x-direction. The force function is F = x^{2} N. What is the work done in moving the object from x = 2 to 5?

a) 40

b) 39

c) 38

d) 69

View Answer

Explanation: The work is given by formula W = \(\int_{x1}^{x2}\)F * dx. So, W = \(\int_{2}^{5}\)x

^{2}* dx = (5

^{3}-2

^{3})/3 = 39 J.

10. A particle is acted upon by a force in x direction which varies as shown in the graph. What is the work done on the particle?

a) 100 J

b) 10 J

c) 200 J

d) 150 J

View Answer

Explanation: The work done is the area under the force versus displacement graph. The area can be calculated as area of trapezium i.e. 0.5*(sum of parallel sides)*height = 0.5*(6+4)*2 = 10.

11. What is the SI unit of variable force?

a) Newtons

b) Joules

c) Ampere

d) Mole

View Answer

Explanation: The SI unit of force is Newtons. While Joules, Ampere, and Mole are SI units of energy, current and amount of substance.

12. A spring is stretched by a length x, then the potential energy stored in it changes by 0.5*k*x^{2}, where k is the spring constant.

a) True

b) False

View Answer

Explanation: The potential energy of a stretched spring is given by 0.5*k*x

^{2}where x is the elongation in the spring.

13. A force of 600 N elongates a spring from its natural length of 18 cm to a length of 20 cm. What is the work done?

a) 116J

b) 114J

c) 118J

d) 120J

View Answer

Explanation: The force on a spring stretched by x meters is given by F = k*x where k is the spring constant. So, 600 = k*(2/100). Now k = 3*10

^{4}, and hence F = 30000*x. W = \(\int_{x1}^{x2}\)F * dx = \(\int_{0.18}^{0.2}\)30000 * x * dx = 114 J.

14. Which of the following is not a variable force?

a) F = x

b) F = y^{2}

c) F = 2

d) F = 2*t

View Answer

Explanation: F = x, F = x

^{2}varies with x and y co-ordinate respectively. F = 2*t varies with time but F = 2 is always constant.

15. Which of the following is not a conservative force?

a) Gravitational force

b) Spring force

c) Friction force

d) Magnetic force

View Answer

Explanation: Friction force causes energy losses in form of heat so it is a non-conservative force while gravitational, spring and magnetic forces are all conservative.

16. A force F = x + 1 N acts on a particle in x direction. What is the work done from x = 2m to 5 m?

a) 10.5 J

b) 14.5 J

c) 13.5 J

d) 0.5 J

View Answer

Explanation: Work done is given by W = \(\int_{x1}^{x2}\)F * dx. So, W = \(\int_{2}^{5}\)(x + 1) * dx = [(5

^{2}/2 + 5) – (2

^{2}/2+2)] = 13.5.

17. What is the value of spring constant if force of 200 N is acting on it and the spring is compressed by 2m?

a) 100 kgs^{-2}

b) 100 kgs^{-1}

c) 100 kg^{2}s^{2}

d) 100 Kgs

View Answer

Explanation: Force on a elongated/compressed spring is given by F = k*x where k is spring constant and x is elongation/compression. So, 200 = k*2 which gives k = 100 kgs

^{-2}.

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