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# Work, Energy and Power

#### Multiple Choice Questions

11.

• 100J

• 50J

• 200J

• 75J

A.

100J

12.

The work done in pulling up a block of wood weighing 2 kN for a length of 10 m on a smooth plane inclined at an angle of 15° with the horizontal is

• 9.82 kJ

• 89 kJ

• 4.35 kJ

• 5.17 kJ

D.

5.17 kJ

Here :- Weight of block;ω = 2kN; Distance d =10m

Angle of inclination on the plane α =150

The block will be pulled up on a smooth plane

Hence,force of resistance due to the inclination

F =ωsinα = 2×103×Sin 150

=2×103×0.2588

=0.5176 kN

Now work done W = force×displacement =0.1576×103×10=5.17kJ

13.

The spring extends by x on loading, then energy stored by the spring is (if T is the tension in spring and k is
spring constant

• $\frac{{\mathrm{T}}^{2}}{2\mathrm{k}}$

• $\frac{{\mathrm{T}}^{2}}{2{\mathrm{k}}^{2}}$

• $\frac{2k}{{\mathrm{T}}^{2}}$

• $\frac{2{\mathrm{T}}^{2}}{k}$

A.

$\frac{{\mathrm{T}}^{2}}{2\mathrm{k}}$

The spring force is an example of variable force which is conservative.

14.

A solid sphere is rolling on a frictionless surface, show in figure with a translational velocity v m/s. If it is to climb the inclined surface then v should be

• ≥ $\sqrt{10}{7\mathrm{gh}}}$

• ≥ $\sqrt{2\mathrm{gh}}$

• 2gh

• 10/7 gh

A.

≥ $\sqrt{10}{7\mathrm{gh}}}$

From conservation of energy

Potential energy = translational kinetic energy + rotational kinetic energy

mgh = 2

mgh  =

⇒  = mgh

v2 =$\frac{10}{7}$gh

⇒    v = $\sqrt{\frac{10}{7}}$ gh

Hence too climb the inclined surface v should be  ≥ $\sqrt{\frac{10}{7}\mathrm{gh}}$

15.

If the heat of 110 J is added to a gaseous system, whose internal energy is 40 J, then the amount of external work done is

• 180J

• 70J

• 110J

• 30J

B.

70J

Heat added to the system ΔQ =110J

Internal energy Δu =40J

External work done is given by

ΔW= ΔQ-Δu=110-40=70J

16.

A bomb of mass 3.0 kg in air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at speed of 80 m/s. The total energy imparted to the two fragments is

• 1.07 kJ

• 2.14 kJ

• 2.4 kJ

• 4.8 kJ

D.

4.8 kJ

The velocity of second piece can be find out using conservation of momentum.

0 + m1 v1 + m2 v2 = 2 × v1 + 1 × 80

⇒         v1 = - 40 m/s

Negative sign showing that particle is moving in opposite direction of othr particle direction.

The energy imparted to the fragments are converted into their kinetic energy.

∴   Total energy =

=

= 4800 J

Total energy = 4.8 kJ

17.

A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?

• 2 sec

• 4 sec

• 1 sec

• 6 sec

B.

4 sec

When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by

Hence,

18.

A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass 'm' is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period

• decreases continuously

• decreases initially and increases again

• remains unaltered

• increases continuously

B.

decreases initially and increases again

According to law of conservation of momentum

Iω = consant

I = moment of inertia

ω = angular speed

When viscous fluid of mass m is dropped and start spreading out then its moment of inertia increases and angular velocity decreases. But when it start falling then its moment of inertia again starts decreasing and angular velocity increases.

19.

Amount of work which can be obtained from 200 cal heat, will be

• 280 J

• 800 J

• 420 J

• 840 J

D.

840 J

Amount of work done is given by

W=200cal= 200×4.2 J= 840 J       ($\because$1 cal=4.2J)

20.

A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/sec. It is subjected to a retarding force F = -0.1 x joule/meter during its travel from x = 20 meter to x = 30 meter. Its final kinetic energy will be

• 475 joule

• 450 joule

• 275 joule

• 250 joule

A.

475 joule

According to work-energy theorem,

W = Δ K.E

⇒   W =

⇒   F · dx  = K.Ef

⇒   F · dx = K.Ef  $-$ 500

⇒    = K.Ef $-$ 500

⇒   = K.Ef $-$ 500

⇒    = K.Ef $-$ 500

⇒   K.Ef = 500 $-$ 25

⇒    K.Ef = 475 J