Physics Test

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Physics Test - 25

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Question 1
1 / -0

A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.

A
-2.94 cm

B
-3.12 cm

C
2.24 cm

D
1.52 cm

Solution

Here, A = 5 cm ; f = 5 Hz ; t = 0.32 s

Now,y = A sin ωt = A sin 2π f t = 5 sin (576°) = –2.94 cm
Question 2
1 / -0

Two cylindrical rods of uniform cross-sectional area A and 2A, having free electrons per unit volume 2n and n respectively are joined in series. A constant current I flows through them in steady state. The ratio of drift velocity of free electrons in the left rod to that of the right rod is (VL/VR) is:

A
1/2

B
1

C
2

D
4

Solution

Since current I = neAV_d through both rods is same

2(n) e A V_L= n e (2A) V_R

or V_L/V_R=1
Question 3
1 / -0

A tuning fork vibrates with a frequency of 256 Hz. Taking the speed of sound to be 345.6 m s-1 in the air, find the wavelength and the distance, which the sound travels during the time, fork makes 60 vibrations.

A
1.35 m, 81 m

B
1.40 m, 84 m

C
1.45 m, 87 m

D
1.55 m, 93 m

Solution

Here, \(f=256 H z ; v=345.6 \mathrm{ms}^{-1}\)
\(\therefore \lambda=\frac{v}{f}=\frac{345.6}{256}=1.35 \mathrm{m}\)
In one cycle, sound advances by distance \(\lambda\)
Therefore, the required distance travelled by sound,
\(S=60 . \lambda=81 \mathrm{m}\)
Question 4
1 / -0

For network shown in figure, determine V₀and I_D. (Where, I_D = Current flowing in the diodes)

A
11 V; 1.96 mA

B
16 V; 2.32 mA

C
21 V; 2.84 mA

D
28 V; 3.49 mA

Solution

Direction of current I \(p\) is the same direction as arrow in the diode symbol. Therefore, both the diodes are in forward Bias; hence voltage drops across their terminals are \(V_{S i}\) and \(V_{G e}\) are \(0.7 V\) and \(0.3 V\) respectively. \(E=12 V>(0.7 V+0.3 V)\)
\(V_{0}-E-V_{S i}-V_{G e}-12-0.7-0.3-11 V\)
and \(\quad I_{D}=\frac{V_{0}}{R}=\frac{11}{5.6}=1.96 m A\)
Hence, the correct option is (A)
Question 5
1 / -0

Water is filled up to a height 'h' in a cylindrical vessel. Now, a hole of area 'A' is made at bottom of the vessel. Water drains out of the hole in time 't' seconds. If we repeat the above process with a height of water as '4h', then how much time it require for water to drain out of the cylinder?

[Assume A << A₀ (area of tank)]

A
t seconds

B
4t seconds

C
2t seconds

D
t/4 seconds

Solution

Time required to empty the tank
\(t=\frac{A}{A_{0}} \sqrt{\frac{2 H}{g}} \Rightarrow \frac{t_{2}}{t_{1}}=\sqrt{\frac{H_{2}}{H_{1}}}=\sqrt{\frac{8 h}{2 h}}=2\)
\(\therefore \mathrm{t}_{2}=2 \mathrm{t}\) seconds
Hence, the correct option is (C)
Question 6
1 / -0

If the radius of the earth were to shrink by one percent and its mass remains the same, the acceleration due to gravity on the earth's surface would

A
Decrease

B
Remain unchanged

C
Increase

D
Become zero

Solution

g=GM/R²

or g∝1/R²

g will increase if R decreases.
Hence, the correct option is (C)
Question 7
1 / -0

Which of the following statement is false for electromagnetic waves ?

A
Both electric and magnetic field vectors attain the maxima and minima at the same place and at the same time.

B
The energy in electromagnetic wave is divided equally between electric and magnetic fields.

C
Both electric and magnetic field vectors are perpendicular to each other and perpendicular to the direction of propagation of wave.

D
These waves cannot propagate in a material medium.

Solution

Electromagnetic waves can propagate in material as well as vacuum.
Hence, the correct option is (D)
Question 8
1 / -0

A ball is projected up an incline of 30o with a velocity of 30 m s⁻¹at an angle of 30o with reference to the inclined plane from the bottom of the inclined plane. If g=10 m s⁻², then the range on the inclined plane is

A
12 m

B
60 m

C
120 m

D
600 m

Solution

Range on incline plane \(R=\frac{2 u^{2} \sin (\alpha-\beta) \cos \alpha}{g \cos ^{2} \beta}\)
here \(\beta=30^{\circ}\)
\(\alpha=30^{\circ}+30^{\circ}=60^{\circ}\)
\(R=\frac{2(30)^{2} \sin \left(30^{\circ}\right) \cos \left(60^{\circ}\right)}{10 \cos ^{2}\left(30^{\circ}\right)}\)
\(=\frac{2(30)^{2}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}{10\left(\frac{3}{4}\right)}\)
\(=60 \mathrm{m}\)
Hence, the correct option is (B)
Question 9
1 / -0

Which of the following graphs correctly represent the variation of β=−[(dV/dp)/V] with p for an ideal gas at constant temperature?

A

B

C

D

Solution

\(\beta=-\frac{d V / d P}{V}=\) compressibility of gas
\(=\frac{1}{\text { Bulk modulus of elasticity }}\)
and \(\beta=\frac{1}{v}\) under isothermal conditions.
Hence, the correct option is (A)
Question 10
1 / -0

In an NPN transistor, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, then the emitter current will be

A
9 mA

B
11 mA

C
1 mA

D
0.1 mA

Solution

In an NPN transistor emitter current (i_e) is the sum of base current (i_b) and collector current (i_c).

Given, \(\frac{90}{100} i_{e}=i_{c} \ldots\) (i)
Also, \(i_{c}=10 \mathrm{mA} \ldots\) (ii)
From Eqs. (i) and (ii), we get
\(i_{e}=\frac{10 \times 100}{90}=11 \mathrm{mA}\)
Hence, the correct option is (B)