Tag: physics

Questions Related to physics

The period of a particle it is $8s$. At $t=0$ it is at the mean position. The ratio of the distance covered by the particle in first second and second will be

  1. $\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } $

  2. $\cfrac { 1 }{ \sqrt { 2 } } $

  3. $\cfrac { 1 }{ \sqrt { 2 } -1 } $

  4. $\left[ \sqrt { 2 } -1 \right] $


Correct Option: C
Explanation:

$x=A\sin\omega t$

$T= 8 sec$
$\dfrac{2\pi}{\omega}=8 sec$
$\omega=\dfrac{2\pi}{8}=\dfrac{\pi}{4}$
$x=A\sin \dfrac{\pi}{4} t$ ,At $t=1 sec$
$x _1=\dfrac{A}{\sqrt2}$ ,At $t=2sec$
$x _2=A$
Required ration $=\dfrac{x _1}{x _2-x _1}=\dfrac{1/\sqrt2}{1-\dfrac{1}{\sqrt2}}=\dfrac{1}{\sqrt2-1}$

Sunlight of intensity 1.3 kW m–2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, kW m–2, at a distance 22 cm from the lens on the other side is _________. 

  1. $ 130 $

  2. $120 \ $

  3. $ 170$

  4. $160 $


Correct Option: A,C

Simple harmonic motion is the projection of uniform circular motion on the

  1. $x$- axis

  2. $y$- axis

  3. reference circle

  4. any diameter of reference circle.


Correct Option: D
Explanation:

Simple harmonic motion is the projection of uniform circular motion on any diameter of reference circle.

Which one of the following equations of motion represents simple harmonic motion?

  1. Acceleration =$-k _0x+k _1x^2$

  2. Acceleration =$-k(x+a)$

  3. Acceleration =$k(x+a)$

  4. Acceleration =$kx$


Correct Option: B

The mass of particle executing S.H.M is 1 gm.If its periodic time is $\pi $ seconds, the value of force constant is:-

  1. 4 dynes/cm

  2. 4 N/cm

  3. 4 N/m

  4. 4 dynes/m


Correct Option: A
Explanation:

Given, mass $m=1 gm$; time period $T=\pi$ sec

The angular frequency $\omega=\frac{2\pi}{T}=\frac{2\pi}{\pi}=2$
As $\omega=\sqrt{k/m}$  where ($k$ is the force constant),
$k=m\omega^2=(1 gm)(2)^2=4$ dynes/cm

A particle of mass 1kg is moving in a S.H.M with an amplitude of 0.02m and a frequency of 60Hz. The maximum force acting on the particle is :

  1. $2.88 \times 10^{3}$N

  2. $1.44 \times 10^{3}$N

  3. $5.67 \times 10^{3}$N

  4. $0.75 \times 10^{3}$N


Correct Option: A
Explanation:

$f = 60\ Hz$
$w = 2\pi f = 120\pi \  rad/sec$
$F = mw^{2}A$
$= 1 \ kg \times 14400 \times \pi ^{2} \times  0.02$
$= 2.88 \times 10^{3}N$

In case of force oscillations of a body

  1. driving force is constant throughout.

  2. driving force is to be applied only momentarily.

  3. driving force has to be periodic and continuous.

  4. driving force is not required.


Correct Option: C
Explanation:

In case of force oscillations of a body driving force has to be periodic and continuous.

A student says that he had applied a force $F= -k\sqrt {x}$ on a particle and the particle moves in simple harmonic motion. He refuses to tell whether $k$ is a constant or not. Assume that he has worked only with positive $x$ and no other force acted on the particle.

  1. As $x$ increases $k$ increases

  2. As $x$ increases $k$ decreases

  3. As $x$ increases $k$ remains constant

  4. The motion cannot be simple harmonic


Correct Option: A
Explanation:
Given that the motion of the particle is Simple Harmonic Motion.

$F=-m\omega^2 x$

The force applied by the student is given by, 

$F=-\dfrac{-k}{x}$

If the motion is S.H.M., then both will be equal.

So,

$\dfrac{-k}{-x}=m\omega^2 x$

$k=m\omega^2 x^2$

This means, if $x$ increases, then $k$ also increases.

So the correct answer is option a.

The force of a required to row a boat at velocity is proportional to square of its speed of v km/ h requires 4 KW, how many does a seepd of 2V km/h required 

  1. 8 kW

  2. 16 kW

  3. 32kw

  4. 76kw


Correct Option: C
Explanation:
Here $F\propto V^{2}$
$P = F\times V$
$\propto V^{2}\times V = V^{3}$
$\therefore \dfrac{P _{1}}{P _{2}}=\dfrac{V _{1}^{3}}{V _{2}^{3}}=\left(\dfrac{V}{2V}\right)^{3}=\dfrac{1}{8}$
$\therefore P _{1}=4kw$
$P _{2}=8\times 4= 32kw$
$\therefore P _{2}= 32kw$

A body is executing SHM. If the force acting on the body is 6N when the displacement is 2 cm, then the force acting on the body when the displacement is 3 cm in newton is:

  1. $ 6 $ N

  2. $9$ N

  3. $4$ N

  4. $\sqrt{6} $ N


Correct Option: B
Explanation:
$ F = m w^{2}x$
$ \dfrac{F _{1}}{F _{2}} = \dfrac{x _{1}}{x _{2}}$
$ F _{2}  = \dfrac{3}{2}\times 6 \ N  =  9 \ N$