Physics Test

Result

Physics Test - 30

Score
-
out of -
Rank
-
out of -

TIME Taken - -

Weekly Quiz Competition

Question 1
3 / -1

The depletion layer in a \(p-n\) junction diode is \(10^{-6}m\) wide and its knee potential is \(0.5 V\). What is the inner electric field in the depletion region?

A

\(5 \times(10)^{6} \frac{{V}}{{m}}\)

B
\(5 \times(10)^{-7} \frac{{V}}{{m}}\)

C
\(5 \times(10)^{5} \frac{V}{m}\)

D
None of these

Solution

In both forward biasing and reverse biasing, applied potential establishes an internal electric field, which acts against or towards the potential barrier. This internal electric field is weakened or strengthened at the junction. In forward voltage at which the current through the junction starts to increase rapidly, once the applied forward voltage exceeds the knee voltage, the current starts increasing rapidly. In forward biasing condition, the inner electric field is given by:
\({E}=\frac{\Delta{V}}{\Delta {r}}\)
or \(|{E}|=\frac{\Delta{V}}{\Delta {r}}=\frac{5 \times 10^{-1}}{10^{-6}}\)
\(=5 \times 10^{5} {V} / {m}\)
Question 2
3 / -1

A train of length \(80 m\) is moving with a uniform velocity of \(25 m/sec\). The time taken by the train to cross a bridge \(120 m\) long is

A
3.2 sec

B
4.8 sec

C
8 sec

D
None of these

Solution

Time taken by train to cross the bridge \(=\) (length of train \(+\) length of bridge)/ speed of the train.
Or \(200/ 25\) or \(8 sec\) is the answer.
Question 3
3 / -1

A shell at rest, at the origin explodes into three fragments of masses 1 kg, 2 kg and m kg. The pieces weighing 1 kg and 2 kg fly off with a speed of 5 ms–1along x– axis and 6 ms–1along y– axis respectively. If the 'm' kg piece flies off with a speed of 6.5 ms–1, what must be the total mass of the shell?

A
4 kg

B
5 kg

C
3.5 kg

D
4.5 kg

Solution

0
Question 4
3 / -1

For the given circuit to act as a full-wave rectifier, the AC input should be connectedacross ..i... and ..ii... and the DC output should appear across ...iii... and ...iv....

i, ii, iii and iv, respectively, are

A
B, D, A and C

B
B, A, C and D

C
C, A, B and D

D
C, D, B and A

Solution

For the given circuit which is acting as a full-wave rectifier, the \(AC\) input should be connected across \(B\) and \(D\) and the \(DC\) output should appear across \(A\) and \(C\).

Hence, the correct option is (B).
Question 5
3 / -1

Two electric bulbs rated P1watt V volt, and P2watt V volt are connected in parallel, and applied across V volt. The total power will be

A
\(\left(P_{1}+P_{2}\right)\) watt

B
\(\sqrt{P_{1} P_{2}}\) watt

C

\(\frac{P_{1}+P_{2}}{P_{1}+P_{2}} \) watt

D

\(\frac{P_{1}+P_{2}}{P_{1} P_{2}}\) watt

Solution

Let \({R}_{1}\) and \({R}_{2}\) are connected in parallel.
Power across \({R}_{1}\) is \(P_{1}=\frac{V^{2}}{R_{1}}\)
Power across \({R}_{2}\) is \(P_{2}=\frac{V^{2}}{R_{2}}\)
Total power across the system
\(P=\frac{V^{2}}{R} \Rightarrow P=V^{2}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) \Rightarrow P=P_{1}+P_{2}\)
Where
\(\frac{1}{R}=\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\)
Question 6
3 / -1

The Thermo emf produced in a thermocouple is 3μV/oC. If the temperature of the cold junction is 20oC and the Thermo emf is 0.3 mV, the temperature of the hot junction is-

A
80^oC

B
100^oC

C
120^oC

D
140^oC

Solution

We know that,
EMF produced = EMF produced per unit
Rise in temperature \(\times\) Rise in temperature
\(\Rightarrow 0.3 \times 10^{-3}=3 \times 10^{-6}\left(\mathrm{T}_{\mathrm{H}}-20\right)\)
\(\Rightarrow \mathrm{T}_{\mathrm{H}}-20=\frac{0.3 \times 10^{-3}}{3 \times 10^{-6}}\)
\(\Rightarrow \mathrm{T}_{\mathrm{H}}=120^{\circ} \mathrm{C}\)
Question 7
3 / -1

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity \(\vec u\)and the other from rest with uniform acceleration \(\vec f\). Let \(\alpha\)be the angle between their directions of motion. The relative velocity of the second particle w.r.t. first is least after a time of

A
\(\frac{{u} \sin \alpha}{{f}}\)

B
\(\frac{f \cos \alpha}{u}\)

C
\(u \sin \alpha\)

D

\(\frac{u \cos \alpha}{f}\)

Solution

After \(t\), velocity \(=f\times t\)
\(V_{BA}=\sqrt{f^2t^2+u^2-2fut cos \alpha}\)
For max. and min.\(\frac{{d}}{{dt}}\left({V}_{{BA}}^{2}\right)=2{f}^{2} {t}-2 {fu} \cos \alpha=0\)
\({t}=\frac{{u} \cos }{{f}}\)
Question 8
3 / -1

A nucleus decays by+emission, followed by-emission. If the atomic and mass numbers of the parent nucleus are Z and A, respectively, then the corresponding numbers for the daughter nucleus are

A
Z - 1 and A - 1

B
Z + 1 and A

C
Z - 1 and A

D
Z + 1 and A - 1

Solution

For the positron emission, the mass number decreases by 1 and atomic number remains same. Hence, the corresponding numbers for the daughter nucleus are Z - 1 and A.
Question 9
3 / -1

An engine pumps up 100 kg of water through a height of 10 m in 5s. Given that the efficiency of engine is 60%. If g = 10 ms^-2, the power of the engine is :

A
3.3 kW

B
0.33 kW

C
0.033 kW

D
33 kW

Solution

Power is the rate at which a force does work.
Efficiency of engine, \(\eta=60 \%\)
Thus, power\(=\frac{\text { work/time }}{\eta}=\frac{100}{60} \times \frac{m g h}{t}\)
Given,\(m=100 {kg}, h=10 {m}, t=5 {s}\) and \(g=10 ms^{-2}\)
Hence, power\(=\frac{100}{60} \times \frac{100 \times 10 \times 10}{5}\)
\(=3.3 \times 10^{3} {W}\)
\(=3.3 kW\)
Question 10
3 / -1

Two bodies of masses \(m\) and \(4m\) are moving with equal momentum. The ratio of their kinetic energy is

A
1 : 4

B
4 : 1

C
1 : 1

D
1 : 2

Solution

\(\frac{K E_{1}}{K E_{2}}=\frac{\frac{P_{1}^{2}}{2 m_{1}}}{\frac{P_{2}^{2}}{2 m_{2}}}=\frac{P_{1}^{2} \times 2 m_{2}}{P_{2}^{2} \times 2 m_{1}}=4: 1\)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 30

Result

Physics Test - 30

Score

-

Rank

-

Question 1

\(5 \times(10)^{6} \frac{{V}}{{m}}\)

\(5 \times(10)^{-7} \frac{{V}}{{m}}\)

\(5 \times(10)^{5} \frac{V}{m}\)

None of these

Solution

Question 2

3.2 sec

4.8 sec

8 sec

None of these

Solution

Question 3

4 kg

5 kg

3.5 kg

4.5 kg

Solution

Question 4

B, D, A and C

B, A, C and D

C, A, B and D

C, D, B and A

Solution

Question 5

\(\left(P_{1}+P_{2}\right)\) watt

\(\sqrt{P_{1} P_{2}}\) watt

\(\frac{P_{1}+P_{2}}{P_{1}+P_{2}} \) watt

\(\frac{P_{1}+P_{2}}{P_{1} P_{2}}\) watt

Solution

Question 6

80oC

100oC

120oC

140oC

Solution

Question 7

\(\frac{{u} \sin \alpha}{{f}}\)

\(\frac{f \cos \alpha}{u}\)

\(u \sin \alpha\)

\(\frac{u \cos \alpha}{f}\)

Solution

Question 8

Z - 1 and A - 1

Z + 1 and A

Z - 1 and A

Z + 1 and A - 1

Solution

Question 9

3.3 kW

0.33 kW

0.033 kW

33 kW

Solution

Question 10

1 : 4

4 : 1

1 : 1

1 : 2

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

80^oC

100^oC

120^oC

140^oC