Tag: physics

Questions Related to physics

What does the odometer of an automobile measure?

  1. Speed

  2. Distance

  3. Acceleration

  4. All


Correct Option: B
Explanation:

An odometer or odograph is an instrument that indicates distance traveled by the automobile

A laser signal sent towards the moon returns after t seconds.If c is speed of light, then the distance of the moon from the observer is

  1. ct

  2. $ct^{-1}$

  3. $ct^{-2}$

  4. $\displaystyle \frac{ct}{2}$


Correct Option: D
Explanation:

A laser signal sent towards the moon returns after t seconds.

Time taken to reach moon is $\dfrac{time}{2}=\dfrac{t}{2}$
Therefore $distance=speed*time=c*\dfrac{t}{2}$

Distance is a vector quantity. State whether true or false.

  1. True

  2. False


Correct Option: B
Explanation:

Distance is a scalar quantity. So the given statement is false.

Can a moving object cover zero distance?

  1. Always

  2. Sometimes

  3. Never

  4. Can't be said from given condition


Correct Option: C
Explanation:

A moving object always changes its position. Then, it has move along a path with some length.
So, if the body is moving, it will always cover some path which can never have zero length. Thus distance travelled by a moving object can never be zero.

Given 2x = 4, Where x is distance. Find the Value of X.

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: B
Explanation:

The given equation is  $2x  =4$

Thus  $x  = \dfrac{4}{2} = 2$

State whether given statement is True or False.
Path length is scalar Quantity ?

  1. True

  2. False


Correct Option: A
Explanation:

Path length is distance covered by body in reaching from initial point to final point. It may be different for different paths.

Since path length is distance hence it is scalar quantity, it may be different for two paths with same initial and final points.

The distance of closest approach of an $\alpha $-particle projected towards a nucleus with momentum p is r. What will be the distance of closest approach, when the momentum of projected $\alpha $-particle is 2p?

  1. 2r

  2. 4r

  3. $\dfrac { r }{ 2 } $

  4. $\dfrac { r }{ 4 } $


Correct Option: D
Explanation:

At the distance of closest approach:$KE=PE$

$\dfrac{p^2}{2m}=\dfrac{k 2 eq}{r}$
When the momentum chnages to 2p:
$\dfrac{(2p)^2}{2m}=k\dfrac{2eq}{r'}$
$\dfrac{4p^2}{2m}=k\dfrac{2eq}{r'}$
$\dfrac{2p^2}{m}=k\dfrac{2eq}{r'}$
$r'=k\dfrac{2eq}{p^2} \times \dfrac{m}{2}$
$=\dfrac{r}{2m}\times \dfrac{m}{2}$
$r'=\dfrac{r}{4}$

The height $y$ and horizontal distance $x$ covered by a projectile in a time $t$ seconds are given by the equations $y = 8t - 5t^2$ and $x = 6t$. If $x$ and $y$ are measured in meters, the velocity of projection is:-

  1. $10 \ ms^{-1}$

  2. $6 \ ms^{-1}$

  3. $8 \ ms^{-1}$

  4. $14 \ ms^{-1}$


Correct Option: A
Explanation:

$v _x=\dfrac{dx}{dt}=6$

$v _y=\dfrac{dy}{dt}=8-10t$
$\vec v=v _x\hat{i}+v _y\hat{j}$

$\vec u=\vec v _{t=0}$
$\vec u=6\hat{i}+8\hat{j}$
$u=\sqrt{6^2+8^2}$
$u=10m/s$

Estimate the force exerted by a $75 \ W$ bulb on your finger tip (are $1\ cm^{2}$), when is at a distance of $8 \ cm$ from the bulb. (Assume $100\%$ absorption)

  1. $1 \ N$

  2. $3.1 \times 10^{-10} N$

  3. $4 \times 10^{-6} N$

  4. $2 \times 10^{-4} N$


Correct Option: B
Explanation:

$\begin{array}{l} P=\frac { { \overline { S }  } }{ c } =\frac { { \frac { P }{ { 4\pi { d^{ 2 } } } }  } }{ c } =\frac { { \frac { { 75W } }{ { 4\pi { { \left( { 8.0\times { { 10 }^{ -2 } }m } \right)  }^{ 2 } } } }  } }{ { 3.00\times { { 10 }^{ 8 } }\, m/s } } =3.1\times { 10^{ -6 } }N/{ m^{ 2 } }. \ The\, \, force\, \, is\, \, pressure\, \, times\, \, area, \ we\, \, approximate\, \, the\, \, area\, \, of\, \, a\, \, fingertip\, \, to\, \, be\, \, 1.0\, c{ m^{ 2 } } \ F=PA=\left( { 3.1\times { { 10 }^{ -6 } }N/{ m^{ 2 } } } \right) \left( { 1.0\times { { 10 }^{ -4 } }{ m^{ 2 } } } \right) =3.1\times { 10^{ -10 } }N \end{array}$

Hence,
option $(B)$ is correct answer.

A boy of mass $m$ is at one end of a flat boat of mass $M$ and length $l$ which floats stationary on water. If boy moves to the other end,

  1. The boat moves in opposite direction through a distance $d = \dfrac { m l } { ( M + m ) }$

  2. The displacement of boy with respect to ground is $d ^ { 1 } = \dfrac { - M } { ( M + m ) }$

  3. The displacement of boy with reject to ground is $d^{2}=\dfrac{-M}{(M+m)}$

  4. The displacement of boy with reject to ground is$d^{3}=\dfrac{-m}{(m+M)}$


Correct Option: B