[2 Mark Question Answer]

Question 12 Marks

A simple pendulum is suspended from a rigid support $O$. Its resting position is at mean position $A$. When it pulled to one side and then released, it oscillates from one side to other. In doing so it moves equal distance and equal height on either side as shown in the figure.
Complete the table, using the labels from the figure. The first label is done for you.

S.No	Description	Label
(a)	Potential energy at point A	0 (zero)
(b)	The point(s) having maximum velocity
(c)	The point(s) having minimum velocity
(d)	The point(s) having maximum kinetic energy
(e)	The point(s) having minimum potential energy

Answer

(b) A (c) B and C (d) A (e) B and C

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Question 22 Marks

Adjacent diagram shows a body of mass 5 kg pulled up an inclined plane by a force of 30 N .
(a) Calculate force of gravity acting on body. (Take $g =10 ms^{-2}$ )
(b) Work done by the force in pulling body along the inclined plane.
(c) Work done against the force of gravity.

Answer

(a) Force of gravity acting on the body $=m g=5 kg \times 10 ms^{-2}= 5 0 ~ N$
(b) Work done by the force in pulling the body along the inclined plane
$=\text { Force } \times \text { displacement }=30 N \times 5 m=150 J$
(c) Work done against the force of gravity
$=m g h=50 N \times B C=50 N \times \sqrt{\left(5^2 m^2-4^2 m^2\right)}=50 N \times 3 m=150 J$.

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Question 32 Marks

A compressed spring is held near a small toy car of mass 0.15 kg . On the release of the spring, the toy car moves forward with a velocity of $10 ms^{-1}$. Find the potential energy of the spring.

Answer

Kinetic energy of the toy car $=\frac{1}{2} m V^2$
$=\frac{1}{2} \times 0.15 kg \times\left(10 ms^{-1}\right)^2=0.075 \times 10^2 J=7.5 J$
By law of conservation of energy,
P.E. of spring $= K . E$. of the toy car $=7.5 J$

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Question 42 Marks

Calculate the power of an electric pump in horse power, which can lift $2000 m^3$ of water from a depth of 20 m in 25 minutes. [Take $g=10 ms^{-2}, 1 m^3$ of water $=10^3 kg$ and $1 HP =750 W$ ]

Answer

Volume of water $=2000 m^3$
$\therefore$ Mass of water $= V \times D =2000 m^3 \times 10^3 kgm ^{-3}=2 \times 10^6 kg$
Work done, $W =m g h=2 \times 10^6 kg \times 10 ms^{-2} \times 20 m=4 \times 10^8 J$
Power of electric pump $=\frac{ W }{t}=\frac{4 \times 10^8 J}{25 \times 60 s}=\frac{4 \times 10^8}{1500} W$
Power in $HP =\frac{4 \times 10^8}{1500 \times 750} HP =\frac{3200}{9} HP =355.55 HP$

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Question 52 Marks

State the energy changes taking place in the following cases:
(a) Glowing of a torch bulb
(b) An air gun is loaded and then fired
(c) Water in a dam rotates a turbine coupled to a generator

Answer

(a) Chemical energy of the cell to electrical energy and finally into heat and light energy.
(b) Elastic potential energy to kinetic energy.
(c) Gravitational potential energy to kinetic energy of water, which further changes to mechnical energy of turbine and then in generator, it is converted into electric energy.

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Question 62 Marks

Define work and discuss when the work done is maximum and when it is zero.

Answer

Work is said to be done only when the force applied on a body causes a displacement of the body in the direction of force or any resolved component of the force.
The work is maximum, when displacement is in the direction of applied force.
The work done is minimum when displacement is zero or displacement is normal to the direction of the applied force.

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Question 72 Marks

Define power. State two mathematical expressions for power.

Answer

The rate of doing work is called power.
$P=\frac{W}{t}=\frac{\overrightarrow{F} \times \overrightarrow{S}}{t}.........(1)$
$P=\overrightarrow{F} \cdot \vec{V}........(2)$
where,
$W \rightarrow$ Work done; $t \rightarrow$ Time; $V \rightarrow$ Velocity; $F \rightarrow$ Force

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Physics [2 Mark Question Answer]

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