Work Power And Energy(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

WORK POWER AND ENERGY MCQs

Work And Energy, Sources Of Energy, Electric Potential Energy Potential And Dipoles

Total Questions : 109 | Page 7 of 11 pages

Question 61. What is the angle between the following vectors?

\to A = 3^i - 2^j +^k

\to B = 2^i + 6^j - 6^k

cos−1(12√266)
cos−1(−6√266)
cos−1(−6√1064)
None of these

Discuss Question

Answer: Option B. -> cos−1(−6√266)
:
B

\to A . \to B = (3 \times 2) + (- 2 \times 6) + (1 \times - 6)

= 6 - 12 - 6
= -12

| \to A | = \sqrt{3^{2} + (- 2)^{2} + 1^{2}} = \sqrt{9 + 4 + 1} = \sqrt{14}

| \to B | = \sqrt{2^{2} + 6^{2} + (- 6)^{2}} = \sqrt{4 + 36 + 36} = \sqrt{76}

\frac{\to A . \to B}{| \to A | | \to B |} = \frac{- 12}{\sqrt{14} \sqrt{76}} = \frac{- 6}{\sqrt{266}}

∴ c o s θ = \frac{- 6}{\sqrt{266}}

\Rightarrow θ = c o s^{- 1} (\frac{- 6}{\sqrt{266}})

Question 62. A block of mass 5.0 kg slides down from the top of an inclined plane of length 3 m. The first 1 m of the plane is smooth and the next 2 m is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at

30^{\circ}

with the horizontal, find the coefficient of friction on the rough portion.

A Block Of Mass 5.0 Kg Slides Down From The Top Of An Inclin...

2√3
√32
√34
√35

Discuss Question

Answer: Option B. -> √32
:
B
Force Method:
Form P to Q:

v^{2} = O^{2} + 2 a_{1} S_{1}

Where

a_{1} = g s i n 30^{\circ}, s_{1} = 1 m

And form Q to R:

O^{2} = v^{2} + 2 a_{2} s_{2}

Where

a_{2} = g s i n 30^{\circ} - μ g c o s 30^{\circ}, s_{2} = 2 m

Solve to get

μ = \frac{\sqrt{3}}{2}

Question 63. A light and a heavy body have equal kinetic energy. Which one has a greater momentum ?

The light body
The heavy body
Both have equal momentum
It is not possible to say anything without additional information

Discuss Question

Answer: Option B. -> The heavy body
:
B

P = \sqrt{2 m E} i f E a r e e q u a l t h e n P \infty \sqrt{m}

i.e. heavier body will possess greater momentum.

Question 64. A small block of mass 2 kg is kept on a rough inclined surface of inclination

θ = 30^{\circ}

fixed in a lift. The lift goes up with a uniform speed of 1 m

s^{- 1}

and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of 2 s is.

Zero
9.8 J
29.4 J
16.9 J

Discuss Question

Answer: Option B. -> 9.8 J
:
B
Since the lift's speed is constant, in the absence of acceleration, there will be no pseudo-force (referring to the lift as the frame of reference) or additional reaction/thrust due to inclined plane (referring to ground as the reference frame) on the particle.
Therefore, friction =mg sin

θ

acting along the plane.
Distance moved by the particle (or lift) in time t=vt
Work done in time t=(mg sin

θ

)vt(cos

(90^{\circ}

θ

))=mg

s i n^{2} θ

A Small Block Of Mass 2 Kg Is Kept On A Rough Inclined Surfa...

Substituting the Values we get Work = 9.8 J

Question 65. A 50 g bullet moving with velocity 10 m/s strikes a block of mass 950 g at rest and gets embedded in it. The loss in kinetic energy will be

100%
95%
5%
50%

Discuss Question

Answer: Option B. -> 95%
:
B

A 50 G Bullet Moving With Velocity 10 M/s Strikes A Block Of...

Initial K.E. of system = K.E. of the bullet =

\frac{1}{2} m_{s} v_{s}^{2}

By the law of conservation of linear momentum

m_{s} v_{s} + 0 = m_{s y s} \times v_{s y s}

\Rightarrow v_{s y s} = \frac{m_{s} v_{s}}{m_{s y s}} = \frac{50 \times 10}{50 + 950} = 0.5 m / s

Fractional loss in K.E. =

\frac{\frac{1}{2} m_{s} v_{s}^{2} - \frac{1}{2} m_{s y s} v_{s y s}^{2}}{\frac{1}{2} m_{s} v_{s}^{2}}

By substituting

m_{s} = 50 \times 10^{- 3} k g, v_{s} = 10 m / s

m_{s y s} = 1 k g, v_{s y s} = 0.5 m / s

we get
Fractional loss =

\frac{95}{100}

∴

Percentage loss = 95 %

Question 66.

⃗ F = 3^i + 6^j - 8^k

⃗ s = - 2^i + 5^j + 2^k

If W =

⃗ F . ⃗ s

find W.

8 N. m
20 N. m
5 N. m
None of these

Discuss Question

Answer: Option A. -> 8 N. m
:
A

⃗ F = 3^i + 6^j - 8^k

⃗ s = - 2^i + 5^j + 2^k

W =

⃗ F . ⃗ s

(3^i + 6^j - 8^k) . (- 2^i + 5^j + 2^k)

= (-6) + (30) + (-16)

∵ (^i .^i = 1 &^i .^j = 0)

W = 8 N.m

Question 67. Given

⃗ F = (x y^{2})^i + (x^{2} y)^j N

. The work done by

⃗ F

when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is

Given ⃗F=(xy2)^i+(x2y)^jN. The Work Done By ⃗F When A P...

653J
752J
734J
Zero

Discuss Question

Answer: Option D. -> Zero
:
D
Given

⃗ F = (x y^{2})^i + (x^{2} y)^j

W =

\int F_{x} d x + \int F_{y} d y

\int x y^{2} d x + \int x^{2} y d y

\frac{1}{2} \int d (x^{2} y^{2}) = {[\frac{x^{2} y^{2}}{2}]}_{(0, 0)}^{(4, 0)} = 0

Question 68. The main constituent of biogas is

methane
carbon dioxide
hydrogen
hydrogen sulphide

Discuss Question

Answer: Option A. -> methane
:
A
Bio-gas is an excellent fuel as it contains up to 75% methane. It burns without smoke. Unlike wood, charcoal or coal itleaves no residue like ash.

Question 69. Floating turbine generators are used in the sea to harness ______

hydrostatic energy
wave energy
Solar energy
thermal energy

Discuss Question

Answer: Option B. -> wave energy
:
B
The kinetic energy possessed by huge waves near the seashore can be trappedto generate electricity. The waves are generated by strong winds blowing across the sea. A wide variety of devices have been developed to trap wave energy for rotation of turbine and production of electricity.
Hydrostatic energy is the energy possessed by static water i.e. potential energy

Question 70. The molten material mixed with gases in the mantle of the earth is called ______.

core
lava
geyser
magma

Discuss Question

Answer: Option D. -> magma
:
D
The molten material mixed with gases in the mantle of the earth is called magma.It is a hot fluidmaterial found below Earth's crust. When this magma reaches Earth's crust due to any crack in the crust or volcanoes, then we call it lava. Lave on cooling forms igneous rock.