Physics Test

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Physics Test - 5

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Question 1
3 / -1

A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s . The velocity of the image in cm/s at that instant

A
6cm/s, towards the mirror

B
6cm/s, away from the mirror

C
9cm/s, away from the mirror

D
9cm/s, towards the mirror

Solution

Using the mirror formula, \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
Now, \(f=-12\) \(u=-20\)
\(u^{\prime}=4\)
Now, \(v =-30\)
Differentiating the mirror formula,
we get \(\frac{u^{\prime}}{u^{2}}=\frac{-p^{\prime}}{v^{2}}\)
Substituting the values obtained \(v^{\prime}=\frac{-4 \times 900}{400}\)
Hence, the velocity of the image is \(9 cm / sec\) away from the mirror
Hence option C is correct.
Question 2
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct.

Two blocks A and B, each of mass m, are connected by a massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in the figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then,

A
the kinetic energy of the A-B system, at maximum compression of the spring, is zero

B
the kinetic energy of the A-B system, at maximum compression of the spring, is mv²/4

C
the maximum compression of the spring is v √(m/k)

D
None of the above

Solution

As the collision between block \(C\) and \(A\) is perfectly elastic and their masses are same hence velocity will interchange. After collision between \(C\) and \(A, C\) stops while \(A\) moves with speed of \(C\) ie. \(v\) At maximum compression, \(A\) and \(B\) will move with same speed \(v^{\prime}\) From conservation of linear momentum \(m v=m v^{\prime}+m v^{\prime}=2 m v^{\prime}\)

\(v^{\prime}=\frac{v}{2}\)

Let \(x\) be the maximum compression in this position. At the maximum compression the kinetic energy of the two blocks is \(\frac{1}{2} 2 m\left(\frac{v}{2}\right)^{2}=\frac{1}{4} m v^{2}\)

Using conservation of mechanical energy, we have \(\frac{1}{2} m v^{2}=\frac{1}{2} k x^{2}+\frac{1}{4} m v^{2}\)

\(\frac{1}{2} k \alpha^{2}=\frac{1}{4} m v^{2} \Rightarrow x=v \sqrt{\frac{m}{2 k}}\)

Hence option B is the correct answer.
Question 3
3 / -1

Two blocks P and Q are connected by a light inextensible string passing over a smooth pulley fixed as shown in the Figure. The coefficient of friction of blocks P and Q to the table is μ = 0.3; the mass of the block Q is 20 kg. The mass of the block P for the block just to slip is

A
48.6 kg

B
50.6 kg

C
52.4 kg

D
54.8 kg

Solution

F.B.D. of \(Q\) From F.B.D of \(P\) \(N+T \sin \theta=m g \ldots(1)\)
\(T \cos \theta=f_{\max }=\mu N \ldots(2)\)
From \((1), N=m g-T \sin \theta \ldots(3)\)
From (2) and (3) \(T \cos \theta=\mu(m g-T \sin \theta)\)
\(T(\cos \theta+\mu \sin \theta)=\mu m g \ldots(4)\)
From F.B.D. of \(Q\) \(T=20 g \ldots .(5)\)
\(\frac{(5)}{(4)}\) gives
\(\frac{\mu}{\cos \theta+\mu \sin \theta}=\frac{20}{m}\)
\(\frac{m}{20}=\frac{\cos 60^{\circ}+(0.3) \sin 60^{\circ}}{0.3}\)
\(=\frac{\frac{1}{2}+0.3 \times \frac{\sqrt{3}}{2}}{0.3}\)
\(=\frac{0.5+0.3 \times 0.866}{0.3}\)
\(=2.53\)
\(m=20 \times 2.53=50.6 kg\)
Hence option B is correct.
Question 4
3 / -1

A monochromatic light source of wavelength \(\lambda\) is placed at S. Three slits \(S_{1}, S_{2}\) and \(S_{3}\) are equidistance from the source \(S\) and the point \(P\) on the screen. \(S_{1} P-S_{2} P=\lambda / 6\) and \(S_{1} P-S_{3} P=2 \lambda / 3\). If I be the intensity at \(P\) when only one slit is open, the intensity at \(P\) when all the three slits are open is?

A
3I

B
5I

C
8I

D
zero

Solution

Given, \(S_{1} P-S_{2} P=\frac{\lambda}{6}\)
Hence, phase difference between \(S_{1}\) and \(S_{2}, \phi=\frac{\lambda}{6} \times \frac{2 \pi}{\lambda}=\frac{\pi}{3}\)
So intensity between \(S_{1}\) and \(S_{2}\) is, \(I^{\prime}=I \cos ^{2}\left(\frac{\pi}{6}\right)^{2}=\frac{3 I}{4} \quad \ldots \ldots \ldots(i)\)
Given, \(S_{1} P-S_{3} P=\frac{2 \lambda}{3}\)
Hence, phase difference between \(S_{1}\) and \(S_{3}, \phi=\frac{2 \lambda}{3} \times \frac{2 \pi}{\lambda}=\frac{4 \pi}{3}\)
So intensity between \(S_{1}\) and \(S_{2}\) is \(I^{\prime \prime}=I \cos ^{2}\left(\frac{2 \pi}{3}\right)^{2}=\frac{I}{4} \quad \ldots \ldots \ldots(i i)\)
The intensity at \(P\) when all the three slits are open is, \(I \frac{I^{\prime}}{I^{\prime \prime}}=3 I\)
Hence option A is correct.
Question 5
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct.

The coordinates of a particle moving in a plane are given by:

x(t) = a cos (pt) and y(t) = b sin (pt), where a, b (< a) and p are positive constants of appropriate dimensions.

Which of the following statements is/are correct?

A
The path of the particle is an ellipse.

B
The velocity and acceleration of the particle are normal to each other at (t = π/2p).

C
The acceleration of the particle is always directed towards a focus.

D
All are correct

Solution

Given \(x=a \operatorname{cospt} \quad\) or \(\quad \frac{x}{a}=\operatorname{cosp} t\)
\(y=b \sin p t \quad\) or \(\quad \frac{y}{b}=\sin p t\)
Squaring and adding both we have, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Hence particle will follow an elliptical path Position vector of the particle can be written as \(\vec{r}=a \cos (p t) \hat{i}+b \sin (p t) \hat{j}\)
Differentiating 1 and 2 \(v_{x}=\frac{d x}{d t}=-\operatorname{apsin}(p t) \quad\) and \(\quad v_{y}=\frac{d y}{d t}=b p \cos (p t)\)
Acceleration of the particle \(a_{x}=\frac{d v_{x}}{d t}=-a p^{2} \cos (p t)\)
\(a_{y}=\frac{d v_{y}}{d t}=-b p^{2} \sin (p t)\)
\(a=-a p^{2} \operatorname{cospt}(i)-b p^{2} \sin p t(j\)
Acceleration vector of the particle can be written as \(\vec{a}=-a p^{2} \cos (p t) \hat{i}-b p^{2} \sin (p t) \hat{j}=-p^{2}\left(a \cos (p t) \hat{i}-b p^{2} \sin (p t) \hat{j}\right)\)
\(\vec{a}=-p^{2} \vec{r},\) Hence acceleration is always directed towards the focus At \(t=\frac{\pi}{2 p} \Rightarrow p t=\frac{\pi}{2}\)
\(a_{x}=0\) and \(v_{y}=0 \quad\) and \(\quad v_{x}=-a p \quad\) and \(a_{y}=-b p^{2}\)
So velocity vector is perpendicular to the acceleration vector.

Hence option D is correct.
Question 6
3 / -1

If \(E, M, J\) and \(G\) denote energy, mass, angular momentum and gravitational constant respectively. Then the dimensions of \(\frac{E J^{2}}{M^{5} G^{2}}\) are same as that of:

A
Length

B
Mass

C
Angle

D
Time

Solution

As we know,
\(E=M^{1} L^{2} T^{-2}\)
\(M=M^{1}\)
\(J=M^{1} L^{2} T^{-1}\)
\(G=\left(M^{1} L^{1} T^{-2}\right) \times \frac{L^{2}}{M^{2}}=M^{-1} L^{3} T^{-2}\)
\(\frac{E J^{2}}{M^{5} G^{2}}=\frac{\left(M^{1} L^{2} T^{-2}\right) \times\left(M^{1} L^{2} T^{-1}\right)^{2}}{M^{5} \times\left(M^{-1} L^{3} T^{-2}\right)^{2}}=\frac{M^{3} \times L^{6} T^{-4}}{M^{2} \times L^{6} T^{-4}}=M^{0} L^{0} T^{0}\)
Angle \(=M^{0} L^{0} T^{0}\)
Length \(=M^{0} L^{1} T^{0}\)
Mass \(=M^{1} L^{0} T^{0}\)
Time \(=M^{0} L^{0} T^{1}\)
Hence the correct option is (C).
Question 7
3 / -1

An ammeter is designed by shunting a 60 Ω galvanometer with a 60 Ω resistance. The additional shunt that should be connected across it to double the range is

A
20 Ω

B
25 Ω

C
30 Ω

D
35 Ω

Solution

The shunt for ammeter
\(S =\left\{\frac{ I _{ g }}{ I - I _{ g }}\right\} \times G\)
\(\therefore \frac{G}{S}=\left(\frac{I-I_{g}}{I_{g}}\right)=\left(\frac{1}{1_{g}}-1\right)\)
\(\frac{60}{60}=\frac{1}{1_{8}}-1\) or \(\frac{1}{I_{8}}=2\)
\(\Rightarrow I=2 I_{g}\)
When the range is doubled, \(I=4 I_{g}\) the shunt required Now initial shunt resistance is \(60 .\) Another shunt let have resistance \(R\). It will be connected parallely with first resistance and equivalent resistance become 20.
\(So\)
\(R_{e q}=\frac{60 R}{60+R}\)
\(\Rightarrow 20=\frac{60 R}{60+R}\)
\(\Rightarrow R=30 \Omega\)
Hence option C is correct.
Question 8
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct.

A particle of mass m moves on the x-axis as follows:

It starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times (0 < t < 1). If α denotes the instantaneous acceleration of the particle, then

A
α cannot remain positive for all t in the interval 0 ≤ t ≤ 1

B
|α| cannot exceed 2 at any point in its path

C
α must change sign during the motion, but no other assertion can be made with the information given

D
None of these

Solution

since the body is at rest at \(x=0\) and \(x=1, \alpha\) cannot be positive for all time in the interval \(0 \leq t\) \(\leq 1 .\) Therefore, first the particle is accelerated and then retarded. Now, total time \(t=1 s\) (given) Total displacement \(s=1 m\) (given) \(s=\) Area under \(v\) -t graph
\(\therefore\) Height or \(v_{\max }=\frac{2 s}{t}=2 m / s\) is also fixed. \(\left[\right.\) Area or \(\left.s=\frac{1}{2} \times t \times v_{\max }\right]\)
If height and base are fixed, area is also fixed. In case 2 : Acceleration \(=\) Retardation \(=4 m / s^{2}\) (slope of the graph) In case 1: Acceleration \(>4 m / s^{2},\) while retardation \(<4 m / s^{2}\)

In case 3: Acceleration \(<4 m / s^{2}\) and Retardation \(>4 m / s^{2}\) Hence, \(|\alpha| \geq 4\) at some point \((s)\) in its path.

Hence option A is correct.
Question 9
3 / -1

A circular conducting loop of radius R carries a current I. Another straight infinite conductor carrying current I passes through the diameter of this loop as shown in the figure. The magnitude of force exerted by the straight conductor on the loop is :-

A
πμ0i²

B
μ₀i²

C
(μ₀i²)/2π

D
(μ₀i²)/π

Solution

Take an arc of angle \(d \theta\) which make angle \({ }^{\prime} \theta^{\prime}\) with vertical. \(\left(d \theta\right.\) very small \(\rightarrow\) point \(\left.^{\prime} P^{\prime}\right)\) \(\Rightarrow\) Magnetic field due to infinite wire at this point \(P \Rightarrow B=\) \(\frac{\mu_{0} i}{2 \pi r}\)
(where \(r=R \sin \theta\) i.e. perpendicular distance of 'P' from line) \(\Rightarrow\) force dF on this Point (Arc of length \(d l=R d \theta)\) is \(d F\) \(=i \vec{d} \times \vec{B}\)
\(=i \operatorname{Rd} \theta\left(\frac{\mu_{0} i}{2 \pi R \sin \theta}\right)\)
\(\Rightarrow\) Point (small arc) symmetric to point 'P' will also have same force dF

Force balancing on the arc having angle \(2 \theta\) Total force \(=2 d F \sin \theta\)
\(=2 i R\left(\frac{\mu_{0 i}}{2 \pi R_{\sin \theta}}\right) \sin \theta d \theta\)
\(=\frac{\mu_{0} i^{2}}{\pi} d \theta\)
Force on whole circular loop \(F=\int_{0}^{\pi} \frac{\mu_{0} t^{2}}{\pi} d \theta\)
\(=\frac{\mu_{0} i^{2}}{\pi} \pi\)
\(=\mu_{0} i^{2}\)
Hence option B is correct.
Question 10
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct. A simple pendulum of length \(L\) and mass (bob) \(M\) is oscillating in a plane about a vertical line between angular limits \(-\phi\) and \(+\phi\). For an angular displacement \(\theta(|\theta|<\phi)\), the tension in the string and the velocity of the bob are \(T\) and \(v\), respectively. The relation(s) that hold(s) good under the above conditions is/are:

A
T cosθ = Mg

B
T = Mg cosθ

C
The magnitude of the tangential acceleration of the bob |aT| = g sinθ

D
None of these

Solution

Motion of pendulum is part of a circular motion. In circular motion, resolving of forces occurs in two perpendicular directions; first along radius (towards centre) and second along tangent.

Along radius, Centripetal force, \(T-M g \cos \theta=\frac{m v^{2}}{L}\) Along tangent, Resultant force \(=m a_{T},\) where a \(T\) is the tangential acceleration. \(M g \sin \theta=M a_{T}\)
\(\alpha_{T}=g \sin \theta\)
\(\therefore\) The correct option is c.
Question 11
3 / -1

Directions: The following question has four choices, out of which ONE or MORE can be correct.

A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2 v and v, respectively, strike the bar [as shown in the figure] and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity respectively by ω , E and vc, we have after collision:

A
v_c = 0

B
ω = v/5a

C
E = 3/5 mv²

D
All of the above

Solution

Using conservation of linear momentum,we have

Initial momentum = Final momentum \(p_{i}=m(2 v)-2 m(v)=0 \Rightarrow p_{f}=0,\) hence velocity of the centre of mass of the combined system after collision is zero.

\(v_{c m}=0\)

Using conservation of the angular momentum, we have

Initial angular momentum = Final angular momentum

\(\left(2 m N a+m(2 v)(2 a)=1 \omega \Rightarrow \omega=\frac{6 m v a}{1}\right.\)

Moment of inertia \(I=\frac{1}{12}\left(8 m, 6 a^{2}\right)+2 m a^{2}+m(2 a)^{2}=30 m a^{2}\)

Hence \(\omega=\frac{6 m v a}{3 m m^{2}}=\frac{v}{5 a}\)

Rotational kinetic energy \(K=\frac{1}{2} \mid \omega^{2}=\frac{1}{2}\left(30 \mathrm{ma}^{2}\left(\frac{\mathrm{V}}{5 \mathrm{a}}\right)^{2}=\frac{3}{5} \mathrm{mv}^{2}\right.\)

Hence option D is the correct answer.
Question 12
3 / -1

One million small identical drops of water, all charged to the same potential, are combined to form a single large drop. If E is the sum of the electrostatic energy of each small drop, the combined energy of the large drop is

A
10^–4 E

B
E

C
10⁶ E

D
10⁴ E

Solution

Energy of one drop \(=\frac{1}{2} C V^{2}=\frac{1}{2} q V\) where \(q=C V\) is the charge of one drop, \(C\) is the capacitance Energy of \(10^{6}\) drops, \(E=1 / 2\) qV \(10^{6}\) Potential on the combined drop = (potential of each initial drop) \(\times\left(n^{2 / 3}\right)\) \(V^{\prime}=V\left(10^{6}\right)^{2 / 3}=10^{4} V\)
Energy of combined drop, \(E^{\prime}=\frac{1}{2} q^{\prime} V^{\prime}=\frac{1}{2}\left(10^{6} q\right) V^{\prime}\)
\(=\frac{1}{2} 10^{6} q 10^{4} V\)
\(=10^{4} E\)
Hence option D is correct
Question 13
3 / -1

Consider telecommunication through optical fibres. Which of the following statements is NOT true?

A
Optical fibres can be of graded refractive index

B
Optical fibres are subjected to electromagnetic interference from outside

C
Optical fibres have extremely low transmission loss

D
Optical fibres may have homogeneous core with a suitable cladding

Solution

Some of the characteristics of an optical fibre are as follows:
(i) This works on the principle of total internal reflection.
(ii) It consists of core made up of glass/silica/plastic with refractive index n1, which is surrounded by a glass or plastic cladding with refractive index n2 (n2 > n1). The refractive index of cladding can be either changing abruptly or gradually changing (graded index fibre).
(iii) There is a very little transmission loss through optical fibres.
(iv) There is no interference from stray electric and magnetic field to the signals through optical fibres.
Hence option B is correct.
Question 14
3 / -1

The number of AM broadcast stations that can be accommodated at a 150 kHz band width, if the highest frequency modulating carrier is 5 kHz is

A
15

B
18

C
21

D
24

Solution

Total bandwidth = 150 kHz
f_a(max) = 5 kHz
Any station being modulated by a 5 kHz will produce an upper-side frequency 5 kHz above its carrier and a lower-side frequency 5 kHz below its carrier.
Thus one station needs a bandwidth of 10 kHz.
Number of stations accommodated = Total BW /BW per station
=(150×10³)/(10×10³)=15
Hence option A is correct.
Question 15
3 / -1

A square lead slab of side a = 50 cm and thickness t = 0.5 cm is subjected to a shearing force of magnitude F = 9 × 10⁴ N. If the shear modulus of lead is η = 5.6 × 10⁹ Pa, then how much work is done by the force?

A
5.4 × 10^–3 J

B
6 × 10^–3 J

C
4.5 × 10^–2 J

D
1.44 × 10^–2 J

Solution

\(\eta=\frac{\text { Shearing stress }}{\text { Shearing strain }}\) Work done by the force \(1 / 2 \times\) shear stress \(\times\) shear strain \(\times\) volume \(=\frac{1}{2} \times \frac{F}{a^{2}} \times \frac{F}{a^{2} \eta} \times a^{2} t\)
\(=\frac{F^{2} t}{2 a^{2} \eta}=\frac{\left(9 \times 10^{4}\right)^{2}\left(0.5 \times 10^{-2}\right)}{2(0.5)^{2} \times\left(5.6 \times 10^{9}\right)}\)
\(=1.44 \times 10^{-2} J\)
Hence option D is correct