12 + Interview Questions in PHYSICS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Multiple Correct QuestionThe tube shown has uniform cross-section and is kept on horizontal surface. An ideal liquid is flowing through the tube in the direction represented in the diagram by arrow. Choose the correct statements from the above options.
Answer» Option 2,3 are the correct answer

Discussion

2.	Numerical Based Question
Answer» THE ANSWER TO THIS QUESTION IS 12The particle of mass0.5kg was moving with constant speed parallel to the wall. After being acted upon by an impulse of 2N.s in the direction perpendicular to its motion it would gain a velocity 2/0.5 m/s= 4m/s which would be normal to the wall. After collision with the wall its velocity of recede was 3sqrt2 m/s. Hence coefficient of restitution = (3sqrt2)/4=1(approx)

Discussion

3.	Numerical Based Question
Answer» Let the veloercity of the particle after collision be v As the horizontal components before and after collision remain unchanged then ucosQ = v cos(90-Q)=vsinQ. ....(1) Again the coefficint of restitution e=(vsin(90-Q))/(usinQ)=3/4 .......(2) Comparing (1) and (2) we get (v/u)"2=3/4 =>v= (√3/2)u Option (1) is correct

Discussion

4.	A person of mass M kg is standing on a lift. If the lift moves vertically upwards according to given v-t graph then find out the weight of the man at the following instants :(g=10m/s). Find his apparent weight at t=12 seconds.
Answer» From t=10 sec to t=12 sec s = area under v-t graph = 1/2 * 4 * 20 = 40 mt also s = ut + 1/2 at^2 from solving this you will get a = -5 m/s^2 So the lift is descending with a acc. = 5 m/s^2 So the forces on the person are g downwards ( w.r.t ground ) and a upwards ( w.r.t ground ) so net force on person w.r.t ground is g - a = 10 - 5 = 5 m/s^2 Hence the apparent weight is W = m * a = M * 5 = 5M

Discussion

5.	A diffraction grating has 7,000 lines per centimeter. When the grating is illuminated normally with a monochromatic light, the second order spectral line is found at 62◦. Findi. wavelength of the lightii. third order maximum
Answer» \(d = \frac{10^{-2}}{7000}\) \(\theta = 62° \) \(n = 2\) \(d \,sin\theta_n = n\lambda\) \(\lambda = \frac{d\,sin\theta_n}{n}\) \(\lambda = \frac{10^{-2}}{7000} \times \frac{sin62° }{2}\) \(\lambda = \frac{10^{-2}\times 0.88}{7000\times 2}\) \(\lambda = 0.06285 \times 10^{-5}\) \(\lambda = 6285 \times 10^{-10}\) \(\lambda = 6285 A° \)

5.

A diffraction grating has 7,000 lines per centimeter. When the grating is illuminated normally with a monochromatic light, the second order spectral line is found at 62◦. Findi. wavelength of the lightii. third order maximum

Answer»

\(d = \frac{10^{-2}}{7000}\)

\(\theta = 62° \)

\(n = 2\)

\(d \,sin\theta_n = n\lambda\)

\(\lambda = \frac{d\,sin\theta_n}{n}\)

\(\lambda = \frac{10^{-2}}{7000} \times \frac{sin62° }{2}\)

\(\lambda = \frac{10^{-2}\times 0.88}{7000\times 2}\)

\(\lambda = 0.06285 \times 10^{-5}\)

\(\lambda = 6285 \times 10^{-10}\)

\(\lambda = 6285 A° \)

Discussion

6.	what is matter?
Answer» Matter is a substance that has inertia and occupies physical space. According to modern physics, matter consists of various types of particles, each with mass and size. The most familiar examples of material particles are the electron, the proton and the neutron. Combinations of these particles form atoms.

Discussion

7.	Define the following. 1. compression2. rarefaction3. medium4. vaccum
Answer» (1) Compression: A compression is a region in a longitudinal wave where the particles are closest together. (2) Rarefaction: A rarefaction is a region in a longitudinal wave where the particles are furthest apart. (3) Medium The material or empty space through which signals, waves or forces pass. (4) Vacuum: Vacuum, is a region without any matter, or a region of zero pressure (perfect vacuum). It is practically impossible to create perfect vacuum. So real vacuum should actually be defined as a region almost devoid of matter, or of very low pressure.

7.

Define the following. 1. compression2. rarefaction3. medium4. vaccum

Answer»

(1) Compression:

A compression is a region in a longitudinal wave where the particles are closest together.

(2) Rarefaction:

A rarefaction is a region in a longitudinal wave where the particles are furthest apart.

(3) Medium

The material or empty space through which signals, waves or forces pass.

(4) Vacuum:

Vacuum, is a region without any matter, or a region of zero pressure (perfect vacuum). It is practically impossible to create perfect vacuum. So real vacuum should actually be defined as a region almost devoid of matter, or of very low pressure.

Discussion

8.	Using the method of dimensions find the acceleration of a particle moving with a constant speed v in a circle of radius r.
Answer» \(a \propto v-r\) \(a = kvr\) where a = acceleration v = velocity r = radius \(a = kv^ar^b\) ....(1) \([M^0LT^{-2}] = k[v]^a [r]^b\) \([M^0LT^{-2}] = k[LT^{-1}]^a [L]^b\) \([M^0LT^{-2}] = k[L^{a + b} T^{-a}]\) Comparing power \(a + b = 1\) \(a = 2\) \(b = 1 -2\) \(b= -1\) Then these value put in equation (1) \(a = kv^2r^{-1}\) \(a =\frac{v^2}r\)

8.

Using the method of dimensions find the acceleration of a particle moving with a constant speed v in a circle of radius r.

Answer»

\(a \propto v-r\)

\(a = kvr\)

where

a = acceleration

v = velocity

r = radius

\(a = kv^ar^b\) ....(1)

\([M^0LT^{-2}] = k[v]^a [r]^b\)

\([M^0LT^{-2}] = k[LT^{-1}]^a [L]^b\)

\([M^0LT^{-2}] = k[L^{a + b} T^{-a}]\)

Comparing power

\(a + b = 1\)

\(a = 2\)

\(b = 1 -2\)

\(b= -1\)

Then these value put in equation (1)

\(a = kv^2r^{-1}\)

\(a =\frac{v^2}r\)

Discussion

9.	The period of a simple pendulum is given by T=2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.
Answer» Given \(T = 2\pi \sqrt{\frac lg}\) L.H.S. \([T] = [ M^0L^0T]\) R.H.S. \([l] = [M^0 LT^0]\) \([g] = [M^0 LT^{-2}]\) Then \([T] = \sqrt{\frac lg}\) \([M^0L^0T] = \frac{[M^0LT^0]^{^1/_2}}{[M^0LT^{-2}]^{^1/_2}}\) \([M^0L^0T^{1}] = [M^0L^0T^{1}]\) R.H.S = L.H.S So it is dimensionally correct.

9.

The period of a simple pendulum is given by T=2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.

Answer»

Given \(T = 2\pi \sqrt{\frac lg}\)

L.H.S. \([T] = [ M^0L^0T]\)

R.H.S. \([l] = [M^0 LT^0]\)

\([g] = [M^0 LT^{-2}]\)

Then

\([T] = \sqrt{\frac lg}\)

\([M^0L^0T] = \frac{[M^0LT^0]^{^1/_2}}{[M^0LT^{-2}]^{^1/_2}}\)

\([M^0L^0T^{1}] = [M^0L^0T^{1}]\)

R.H.S = L.H.S

So it is dimensionally correct.

Discussion

10.	The period of a simple pendulum is given by T = 2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.
Answer» Given \(T = 2\pi \sqrt{\frac lg}\) L.H.S. \([T] = [M^0L^0T]\) R.H.S. \([l] = [M^0LT^0]\) \([g] = [M^0LT^{-2}]\) Then \([T] = \sqrt{\frac l g}\) \([M^0L^0T] = \cfrac{[M^0LT^0]^\frac12}{[M^0Lt^{-2}]^\frac12}\) \([M^0L^0T^{-1}] =[M^0L^0T^{-1}]\) R.H.S = L.H.S

10.

The period of a simple pendulum is given by T = 2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.

Answer»

Given \(T = 2\pi \sqrt{\frac lg}\)

L.H.S.

\([T] = [M^0L^0T]\)

R.H.S.

\([l] = [M^0LT^0]\)

\([g] = [M^0LT^{-2}]\)

Then

\([T] = \sqrt{\frac l g}\)

\([M^0L^0T] = \cfrac{[M^0LT^0]^\frac12}{[M^0Lt^{-2}]^\frac12}\)

\([M^0L^0T^{-1}] =[M^0L^0T^{-1}]\)

R.H.S = L.H.S

Discussion

11.	Define SI unit of magnetic field on the basis of Lorentz force
Answer» The SI unit for magnetic field is the Tesla, which can be seen from the magnetic part of the Lorentz force law Fmagnetic = qvB to be composed of (Newton x second)/(Coulomb x meter).

Discussion

12.	∫(7x^5+ 3x^3+ 5/square root of x- 2) dx
Answer» \(\int \left(7x^5 + 3x^3 + \frac5{\sqrt{x - 2}}\right)dx\) \(= \frac{7x^6}{6} + \frac{3x^4}{4} + \frac{5\sqrt{x -2}}{\frac12}+C\) \(\frac76 x^6 +\frac34 x^4 + 10\sqrt{x -2} +C\)

12.

∫(7x^5+ 3x^3+ 5/square root of x- 2) dx

Answer»

\(\int \left(7x^5 + 3x^3 + \frac5{\sqrt{x - 2}}\right)dx\)

\(= \frac{7x^6}{6} + \frac{3x^4}{4} + \frac{5\sqrt{x -2}}{\frac12}+C\)

\(\frac76 x^6 +\frac34 x^4 + 10\sqrt{x -2} +C\)

Discussion

Explore topic-wise InterviewSolutions in .

Multiple Correct QuestionThe tube shown has uniform cross-section and is kept on horizontal surface. An ideal liquid is flowing through the tube in the direction represented in the diagram by arrow. Choose the correct statements from the above options.

Numerical Based Question

Numerical Based Question

A person of mass M kg is standing on a lift. If the lift moves vertically upwards according to given v-t graph then find out the weight of the man at the following instants :(g=10m/s). Find his apparent weight at t=12 seconds.

A diffraction grating has 7,000 lines per centimeter. When the grating is illuminated normally with a monochromatic light, the second order spectral line is found at 62◦. Findi. wavelength of the lightii. third order maximum

what is matter?

Define the following. 1. compression2. rarefaction3. medium4. vaccum

Using the method of dimensions find the acceleration of a particle moving with a constant speed v in a circle of radius r.

The period of a simple pendulum is given by T=2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.

The period of a simple pendulum is given by T = 2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.

Define SI unit of magnetic field on the basis of Lorentz force

∫(7x^5+ 3x^3+ 5/square root of x- 2) dx