Physics Test

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Physics Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

Find the velocity of the point as a function of time and as a function of the distance covered s.

A

\( v_{o} e^{-\frac{s}{R}} \)

B

\(v_{o} e^{-\frac{2 s}{R}} \)

C

\(v_{o} e^{-\frac{s}{2 R}} \)

D

\(2 v_{o} e^{-\frac{s}{R}} \)

Solution

\(a_{t}=-\frac{d v}{d t}\) and \(a_{n}=\frac{v^{2}}{r}\)
\(a_{t}=a_{n}\)
\(-d v / d t=v^{2} / R\)
Also \(\frac{d v}{d t}=\frac{v d v}{d s}\)
From above two equations, \(-d v / v=d s / R\) \(\Rightarrow v=v_{o} e^{-\frac{s}{R}}\)
Hence, the correct option is (A)
Question 2
1 / -0

A neutron decays to a proton and an electron. Find the fraction of energy gone to proton if total energy released is k.mp=1836me.

A
11836

B
18361837

C
18351837

D
11837

Solution

\(0=1836 V 1+v 2\) or \(v l=v 2 I 836\)

total \(K E=I 2[I 836 v 2 I+v 22]\)

\(=12[v 221836+v 22]\)

\(K P K T=12 v 22 \times 1183612 v 22 \times 18371836=I I 837 .\)
Hence, the correct option is (D)
Question 3
1 / -0

For an amplitude modulated wave, the maximum amplitude is found to be 12V and minimum amplitude is found to be 4V. The modulation index of this wave is __________ %.

A
50

B
25

C
75

D
20

Solution

Given, \(Vmax = I 2 V\)
\(Vmin =4 V\)
The modulation index (M) \(=V\) max-VminVmax \(+\) Vmin \(\times 100\)
\(\begin{array}{l}
=12-412+4 \times 100=816 \times 100=80016 \\
=50 \%
\end{array}
\)
Hence, the correct option is (A)
Question 4
1 / -0

A bullet weighing 50 gm leaves the gun with a velocity of 30 ms−1−1. If the recoil speed imparted to the gun is 1 ms−1−1, the mass of the gun is:

A
15 kg

B
1.5kg

C
20kg

D
30kg

Solution

Given, \(mb =50 gm =50 \times 10-3\)
\(vb =3 Oms - I\)
\(vg = Ims - I\)
According to the law of conservation of momentum mgvg=mbvb \(mg = mbvbvg\)
\(mg =(50 \times 10-3)(30) Ig\)
\(mg =150 \times 10-3=1.5 kg\)
Hence, the correct option is (B)
Question 5
1 / -0

Two charges are placed a certain distance apart in air. Ifa glass slab is introduced between them, the force between them will.

A
increase

B
decrease

C
remains the same

D
become zero

Solution

In air, the force between charges is \(F = q 1 q 24 \pi e\) or 2
When the glass slab is introduced between them, the permeability becomes \(\epsilon= K \epsilon O\) where \(K =\) dielectric constant of glass.
Thus, force becomes, \(F ^{\prime}= q Iq 24 \pi KeOr 2\)
Hence, \(F ^{\prime}< F\)
Hence, the correct option is (B)
Question 6
1 / -0

The angle of contact between glass and water is 0o and water (surface tension 70dyn/cm) rises in a glass capillary upto to 6cm. Another liquid of surface tension 140 dyne/cm, angle of contact 600 and relative density 2 will rise in the same capillary up to.

A
12 cm

B
24 cm

C
3 cm

D
6 cm

Solution

The height h through which a liquid will rise in a capillary tube of radius \(r\) is given by \(h=2 S \cos \theta r \mathrm{pg}\)

where \(S\) is the surface tension, \(\rho\) is the density of the liquid and \(\theta\) is the angle of contact. for same capillary and two liquids.

\(h 1 h 2=2\) Sicos \thetairplg \(\times\) rp2 \(g 2 S 2 \cos \theta 2 \Rightarrow h 1 h 2\)

\(=\operatorname{Sicos} \theta \operatorname{Ip} I \times \rho 2 S 2 \cos \theta 2 \Rightarrow 6 h 2=7 O \times 11 \times 2140 \times(1 / 2) \Rightarrow h 2=3 \mathrm{cm}\)
Hence, the correct option is (C)
Question 7
1 / -0

A capillary tube is dipped in water up to length l, the level of water reaches up to height h. Now the end which is inside the water is closed and the capillary tube is put out side the water and that closed end is opened if l > h, the height of the remaining water column in the capillary will be:

A
0

B
l + h

C
2h

D
h

Solution

Liquid rises to height \(h\), which can be obtained by ascent formula as
\(h =2 \operatorname{Tr} \rho g \ldots\)(I)
When the tube is taken out, liquid weight is balanced by upward surface tension forces.
\(\Rightarrow W =2 \pi r T +2 \pi r T =4 \pi r T\) (considering surface tension forces at both \(A\) and \(B )\)
\(\Rightarrow(\pi r 2 H) \rho g=4 \pi r T \Rightarrow H=4 \operatorname{Tr} p g \ldots .(I I)\)
from (I) and (II), we have
\(H =2 h\)
Hence, the correct option is (C)
Question 8
1 / -0

The flux linked with a coil at any instant 't' is given by ϕ=10t2−50t+250. The induced emf at t=3s is:

A
-190 V

B
-10 V

C
10 V

D
190 V

Solution

Farady's law states that time varying magnetic flux can induce an e.m.f. \(E =\) Electric field induced
\(E =- d \phi dt\)
\(E(t)=-d(10 t 2-50 t+250) d t=-(20 t-50) V\)
\(E(t=3 s)=-(20(3)-50) V=-10 V\)
Hence, the correct option is (B)
Question 9
1 / -0

Two parallel large thin metal sheets have equal surface charge

densities (σ=26.4×10−12c/m2) of opposite signs.

The electric field between these sheets is:

A
1.5N/C

B
1.5×10−10N/C

C
3N/C

D
3×10−10N/C

Solution

The electric field between two parallel large plates is \(E =\sigma \in O=26.4 \times 10-128.85 \times 10-12=2.98 \sim 3 N / C\)
Hence, the correct option is (C)
Question 10
1 / -0

A conducting square loop of side ll and resistance R moves in its plane with a uniform velocity vv perpendicular to one of its sides. A uniform and constant magnitude field B exists along the perpendicular to the plane of the loop as shown in figure. The current induced in the loop is:

A
BlvR clockwise

B
BlvR anti-clockwise

C
2BlvR anti-clockwise

D
zero

Solution

Flux at any point in time across the loop = Bl2

the flux does not change with time, hence emf is not induced , hence no current flows.
Hence option D is correct.

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Re-Attempt Weekly Quiz Competition

Physics Test - 12

Result

Physics Test - 12

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SHARING IS CARING

Question 1

\( v_{o} e^{-\frac{s}{R}} \)

\(v_{o} e^{-\frac{2 s}{R}} \)

\(v_{o} e^{-\frac{s}{2 R}} \)

\(2 v_{o} e^{-\frac{s}{R}} \)

Solution

Question 2

11836

18361837

18351837

11837

Solution

Question 3

50

25

75

20

Solution

Question 4

15 kg

1.5kg

20kg

30kg

Solution

Question 5

increase

decrease

remains the same

become zero

Solution

Question 6

12 cm

24 cm

3 cm

6 cm

Solution

Question 7

0

l + h

2h

h

Solution

Question 8

-190 V

-10 V

10 V

190 V

Solution

Question 9

1.5N/C

1.5×10−10N/C

3N/C

3×10−10N/C

Solution

Question 10

BlvR clockwise

BlvR anti-clockwise

2BlvR anti-clockwise

zero

Solution

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