Tag: physics

Questions Related to physics

When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as

  1. simple pendulum

  2. torsional pendulum

  3. compound pendulum

  4. seconds pendulum


Correct Option: C
Explanation:

When a rigid body is suspended vertically, and it oscillates with a small amplitude under the action of the force of gravity, the body is known as compound pendulum. Thus the periodic time of a compound pendulum is minimum when the distance between the point of suspension and the centre of gravity is equal to the radius of gyration of the body about its centre of gravity.

The correct option is (c)

 The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\ cm$ to $8\ cm$ In $40$ seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$, the time In which amplitude of this pendulum will reduce from $10\ cm$ to $5\ cm$ in carbondioxide will be close to (in $5=1.601, \ln { 2 }  2=0.693$)

  1. $231\ s$

  2. $208\ s$

  3. $161\ s$

  4. $142\ s$


Correct Option: C

The time taken for 20 complete oscillations by a seconds pendulum is :

  1. 20 s

  2. 50 s

  3. 40 s

  4. 5 s


Correct Option: C
Explanation:

Time taken for 1 oscillation is 2 s
Time taken for 20 oscillations is $(2 \times 20) = 40  s$

A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at a constant rate. Which of the following statements describes the variation of the time period (T) of the pendulum as the water flows out?

  1. T decreases first and then increases.

  2. T increases first and then decreases.

  3. T increases throughout.

  4. T does not change.


Correct Option: B
Explanation:

$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$
First distance of comfrom suspension point will increase then decrease.

The frequency of a second's pendulum is

  1. 0.5 Hz

  2. 1.0 Hz

  3. 1.5 Hz

  4. none of these


Correct Option: A
Explanation:

Seconds pendulum takes 2 $sec$ to make one complete oscillation. 

Hence, frequency = $\dfrac{1}{2}$ Hz

The frequency of a second's pendulum is :

  1. $0.5\;Hz$

  2. $1.0\;Hz$

  3. $2.0\;Hz$

  4. $5.5\;Hz$


Correct Option: A
Explanation:

A seconds pendulum is a pendulum whose period is precisely two seconds

A simple pendulum with length $L$ and mass $m$ of the bob is oscillating with an amplitude $a$. 

Then the maximum tension in the string is :

  1. $mg$

  2. $mg[1+(\dfrac{a}{L})^{2}]$

  3. $mg[1+\dfrac{a}{2L}]^{2}$

  4. $mg[1+(\dfrac{a}{L})]^{2}$


Correct Option: D

There is a clock which gives correct time at $20^o$C is subjected to $40^o$C. The coefficient of linear expansion of the pendulum is $12\times 10^{-6}$ per $^oC$, how much is gain or loss in time?

  1. $10.3$ sec/day

  2. $19$ sec/day

  3. $5.5$ sec/day

  4. $6.8$ sec/day


Correct Option: A

Find the length of a simple pendulum such that its time period is $2\ s$.

  1. $99.4\ cm$

  2. $89.4\ cm$

  3. $79.4\ cm$

  4. $109.4\ cm$


Correct Option: A
Explanation:

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$


$T^2\, =\, 4\pi ^2\, \displaystyle \frac {L}{g}$
$\Rightarrow\, 2^2\, =\, 4\, \times\, 3.14\, \times\, 3.14\, \times\, \displaystyle \frac {L}{9.8}$
$\Rightarrow\, L\, =\, \displaystyle \frac {4\, \times\, 9.8}{4\, \times\, 3.14\, \times\, 3.14}\, m\, =\, 0.994\, m\, =\, 99.4\, cm$

A desktop toy pendulum swings back and forth once every $1.0 s$. How long is this pendulum?

  1. $0.25\, m$

  2. $0.50\, m$

  3. $0.15\, m$

  4. $0.30\, m$


Correct Option: A
Explanation:

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)


Putting $T = 1$ in eqn. (1), 


We get $1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ $\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)