Physics Test

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Physics Test - 32

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Weekly Quiz Competition

Question 1
3 / -1

1 KWh = _____ J

A
\(3.6 \times 10^{4} \mathrm{~J}\)

B
\(3.6 \times 10^{5} \mathrm{~J}\)

C
\(3.6 \times 10^{6} \mathrm{~J}\)

D
\(3.6 \times 10^{3} \mathrm{~J}\)

Solution

As we know,

\(1 \ \mathrm{hr}=3600\ \mathrm{sec}\)

\(1\) Kilo-watt \(=1000\ \mathrm{Watt}\)

So,

\(1\ \mathrm{KWh}=1000 \mathrm{Watt}-\mathrm{hour}=1000 \times 3600 \mathrm{~W} \cdot \mathrm{s}=3.6 \times 10^{6} \mathrm{~J}\)
Question 2
3 / -1

Of the diodes shown in the following diagrams, which one is reverse biased?

A

B

C

D

Solution

A diode is said to be reverse biased if \(p\)-type semiconductor of p-n junction is at low potential with respect to \(n\)-type semiconductor of \(p-n\) junction. It is so for circuit (3).
Question 3
3 / -1

A cricket ball of mass \(200 gms\) moving with a velocity of \(20\; m/sec\) is brought to rest by a player in \(0.1\; sec\). The average force applied by the player is

A
\(4 \times 10^{3}\) dynes

B
\(4 \times 10^{4}\) dynes

C
\(4 \times 10^{5}\) dynes

D
\(4 \times 10^{6}\) dynes

Solution

\(|F|=\left|\frac{\Delta p}{\Delta t}\right|\)

\(|F|=200 \times \frac{2000}{0.1}=4 \times 10^{6}\) dynes
Question 4
3 / -1

When an electron-positron pair annihilates, how much energy is released?

A
0.8 x 10-13

B
1.6 x 10-13J

C
3.2 x 10^-13 J

D
4.6 x 10-13J

Solution

\(=_1 \beta^{0}+_{-1} \beta^{\circ}=\underset{\gamma-photon}{h v}+\underset{\gamma-photon}{h v}\)
\(E =2 \times 9 \cdot 1 \times 10^{-31} \times\left(3 \times 10^{8}\right)^{2}\)
\(=16-4 \times 10^{-14} \)
\(=1.64 \times 10^{-13} \)
\( =1.6 \times 10^{-13} \mathrm{J}\)
Question 5
3 / -1

For a transistor, the current gain \(\alpha\)is \(0.96\). It is used as an amplifier in a common base circuit with a load resistance of \(4 k \Omega\). If the dynamic resistance of the emitter-base junction is \(48\Omega\), then the voltage gain is

A
40

B
80

C
120

D
160

Solution

Voltage gain\(=\alpha \times \frac{R_{L}}{R_{D}}=0.96 \times \frac{4000}{48}=80\)
Question 6
3 / -1

An electron moves at right angle to a magnetic field of 1.5 x 10^-2T with a speed of 6 x 10⁷m/s. If the specific charge of the electron is 1.7 x 10¹¹C/kg, the radius of the circular path will be :

A
2.9 cm

B
3.9 cm

C
2.35 cm

D
2 cm

Solution

Force due to magnetic field provides the necessary centripetal force.

When electron enters a magnetic field B the force acting is

Since, the electron describes a circular path the centripetal force acting on it is given by

When magnetic field is acting into the plane of paper, the force is downward and the negatively charged particle describes a clockwise circle.
Question 7
3 / -1

If the forces of magnitude 12 Kg wt, 5 Kg wt and 13 Kg wt acting at a point are in equilibrium, then the angle between the two forces is

A
30⁰

B
900

C
600

D
450

Solution

Since\((13)^ 2=(12)^{2}+(5)^{2}\)

So two of them are inclined at 90 to each other.
Question 8
3 / -1

In the circuit given below, an A.C. voltage V = 220 volt (r.m.s.) is applied. The maximum voltage across the capacitor will be

A
\(220 \times \sqrt{2}\) volts

B
\(110 \times \sqrt{2}\) volt

C
\(220\) volts

D
Cannot be determined

Solution

The maximum voltage across the capacitor will be

\(V_{max}=V_{r.m.s}\times \sqrt{2}= 220\times \sqrt{2}\)
Question 9
3 / -1

If \(M_0\)is the mass of an oxygen isotope \(_8O^{17}\) and \(M_p\)and \(M_n\)are the masses of a proton and a neutron, respectively, then what is the nuclear binding energy of the isotope

A

\(\left({M}_{0}-8 {M}\right) {c}^{2}\)

B

\(\left(8 M_{p}+9 M_{n}-M_{0}\right) c^{2}\)

C

\({M}_{{O}} {c}^{2}\)

D

\(\left(M_{0}-17 M_{n}\right) c^{2}\)

Solution

Here, \(Z=8\) and \(A=17,\) therefore,
B.E. \(=\left(8 M_{p}+(17-8) M_{n}-M_{0}\right) c^{2}\)
\(=\left(8 M_{p}+9 M_{n}-M_{0}\right) c^{2}\)
Question 10
3 / -1

P and Q are like parallel forces. If P is moved parallel to itself through a distance x, then the resultant of P and Q moves through a distance:

A

\(\frac{P_{x}}{p+Q}\)

B

\(\frac{P_{x}}{p-Q}\)

C

\(\frac{P_{x}}{p+2Q}\)

D
none of these

Solution

1st case, \(P.AC=Q.CB\).....(1)

2nd case, \({P}\) is shifted by a distance \({AA}^{\prime}={x}\) and as a consequence suppose the resultant \({R}\) shifts from \(C\) to \(C'\) where \({CC'}={y}\).

\(\therefore P.AC=Q.CB\)

or \(P(AC-x+y)=Q(CB-y)\)

or \(P(-x+y)=-Qy\)

\(\because P.A.C=Q.C.B\) by (1)

\(\therefore (P+Q)y=Px\) or \(y=\frac{P_x}{P+Q}\)