Tag: physics

Questions Related to physics

The unit of universal gas constant is

  1. ${Dymes}/{C}$

  2. $Ergs$ ${cm}/{mole} K$

  3. ${Watt}/{K}$

  4. ${Joule}/{mole} K$


Correct Option: D
Explanation:

Gas equation,
$PV=nRT$

$R=\dfrac{PV}{nT}$

$[R]=\left[ \dfrac{PV}{nT}\right]$

$[R]=\dfrac{Joule}{mole \ K}$

Option D

The SI unit of amplitude of oscillation is

  1. cm

  2. m

  3. km

  4. none of these


Correct Option: B
Explanation:

SI unit of amplitude of oscillation is metre  $(m)$

The wavelength of light is expressed in

  1. Metre

  2. Micron

  3. Light year

  4. Angstrom


Correct Option: D
Explanation:

Angstrom $(\dot A)$ unit of length used chiefly in measuring wavelengths of light, equal to $10^{-10}$ metre, or $0.1$ nanometer.  It is used to express wavelengths of visible light, ultraviolet (UV) light, X rays, and gamma rays.

State whether given statement is True or False.
The SI unit of mass is milligram.

  1. True

  2. False


Correct Option: B
Explanation:

SI unit of mass is kilogram i.e.  $Kg$.
Thus the above statement is false.

If the displacement, velocity and acceleration of a particle at time t be x, v and f respectively, then which one is true?

  1. $f=v^3\dfrac{d^2t}{dx^2}$

  2. $f=-v^3\dfrac{d^2t}{dx^2}$

  3. $f=v^2\dfrac{d^2t}{dx^2}$

  4. $f=-v^2\dfrac{d^2t}{dx^2}$


Correct Option: B
Explanation:

We know that the velocity can be written as:
$v=\dfrac{dx}{dt}$

Therefore, $\dfrac1v=\dfrac{dt}{dx}$

Now, expressing $f$ in terms of velocity.

$\dfrac{d^2t}{dx^2}=\dfrac{d\begin{pmatrix}\dfrac{dt}{dx}\end{pmatrix}}{dx}=\dfrac{d\begin{pmatrix}\frac{1}{v}\end{pmatrix}}{dx}=-\dfrac{1}{v^2}\dfrac{dv}{dt}\times\dfrac{1}{v}$
$\Rightarrow\;f=-v^3\dfrac{d^2t}{dx^2}$


A __________ unit does not change with respect to space and time.

  1. derived

  2. standard

  3. regular

  4. large


Correct Option: B
Explanation:

A unit is said to be a standard unit of does not change with respect to space and time.

Which of the following pairs has not the same dimension?

  1. Stress; pressure

  2. Force; surface tension

  3. Impulse; linear momentum

  4. Frequency; angular velocity


Correct Option: B
Explanation:

Surface Tension$ (T) =\dfrac{ Force}{Length}$  . . . . . (1)

Since, $Force = Mass \times Acceleration$

And, $Acceleration = \dfrac{velocity}{ time}= [L T^{-2}]$

$\therefore $ The dimensional formula of force $= M^1L^1T^{-2}$ . . . . (2)

On substituting equation (2) in equation (1) we get,

Surface Tension $(T) = \dfrac{Force}{  Length}$

Or, $T=\dfrac{[M^1L^1T^{-2}]}{[L^{1}]}=M^1T^{-2}$ 

Therefore, surface tension is dimensionally represented as $[M^1T^{-2}]$


Dimension of force $=M^1L^1T^{-2}$

So the dimension of force and surface tension are different.

The force F is given in terms of time t and displacement x by the equation F = A cosBx + C sin Dt. The dimensional formula of D/B is______

  1. $[M^oL^oT^o]$

  2. $[M^oL^o{T}^{-1}]$

  3. $[M^o{L}^{-1}T^o]$

  4. $[M^o{L}^{1} {T}^{-1}]$


Correct Option: D
Explanation:

$F = A\ cosBx + C\ sinDt$

the argument, $\theta $ of cos or sin should be dimensionless.
therefore,
dimension of Bx = $[MLT]$
$[B] [L'] = [MLT]$

$[B] = [M{L}^{0}T]$

Similarly $[D][T'] = [MLT]$

$[D] = [ML{T}^{0}$
dimension of $DB=\dfrac{[ML{T}^{0}]}{[ML^0T]}$
= $[L^1{T}^{-1}]$

Which of the following conversions is correct?

  1. 1 atm = 1.01 x 10$^4$ Pa

  2. 1 mm of Hg = 133 Pa

  3. 1 bar = 10$^7$ Pa

  4. 1 torr = 10$^2$ Pa


Correct Option: B
Explanation:

 The answer is $0.0075006157584566$. We assume you are converting between and pascal. You can view more details on each measurement unit: mm Hg or pascal The SI derived unit for pressure is the pascal. $1 mm Hg =133.322387415 \ pascal$.

Which of the following is correct for the beam which enters the medium ? 

  1. Travel as a cylindrical beam

  2. Diverge

  3. Converge

  4. Diverge near the axis and converge near the periphery


Correct Option: C
Explanation:

Since the refractive index is less at beam boundary, the ray at the edges of the beam move faster compared to the axis of beam. Hence the beam converges.