Physics Test

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Physics Test - 28

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Weekly Quiz Competition

Question 1
4 / -1

A p-n junction diode is connected to a battery of emf \(5.5 \mathrm{V}\) and external resistance of \(5.1 \mathrm{k} \Omega .\) The barrier

potential in the diode is \(0.4 \mathrm{v}\). The current in the circuit is

A
1.08 mA

B
0.08 mA

C
1 mA

D
1 A

Solution

The potential difference across the resistance \(=5.5-0.4=5.1 \mathrm{V}\)
The current through the resistance is I
So, \(V=I R\)
\(I=\frac{V}{R}=\frac{5.1}{5.1 \times 10^{3}}\)
\(I=1 \times 10^{-3}\)
\(I=1 \mathrm{mA}\)
Question 2
4 / -1

The thermo-emf of a thermocouple varies with the temperature θ of the hot junction as E = aθ + bθ² in volts ,where the ratio a/b is 700°C. If the cold junction is kept at 0°C, what is the neutral temperature?

A
700°C

B
350°C

C
1400°C

D
no neutral temperature is possible for this thermocouple

Solution

For the given thermocouple the induced emf increases with the increase in the temperature of the hot junction so no neutral temperature is possible for this thermocouple
Question 3
4 / -1

If the acceleration due to gravity is 'g' on the surface of Earth, then the value of acceleration due to gravity at a height of 32 km above the surface of Earth is (radius of Earth = 6400 km)

A
0.99g

B
0.8g

C
1.01g

D
0.9g

Solution

The acceleration due to gravity at a height h above the surface of Earth is given by

\(g \doteq g \frac{R^{2}}{(R+h)^{2}}\)
For \(h=32 \mathrm{km},\) we can use
\(g^{\prime}=g\left(1-\frac{2 h}{R}\right),\) forh \(<\(g^{\prime}=g\left(1-\frac{2 * 32}{6400}\right)=0.99 g\)
Question 4
4 / -1

The maximum height attained by a projectile is increased by 10% by increasing its speed of projection, without changing the angle of projection. The percentage increase in the horizontal range will be?

A
20%

B
15%

C
10%

D
5%

Solution

We know that \(h=\frac{u^{2} \sin ^{2} \theta}{2 g} .\) The increase \(\delta h\) in \(h\) when \(u\) changes by \(\delta u\) can be obtained by partially differenting this expression. Thus
\(\therefore\)
\[
\begin{aligned}
\delta h &=\frac{2 u \delta u \sin ^{2} \theta}{2 g} \\
\frac{\delta h}{h} &=\frac{2 \delta u}{u}
\end{aligned}
\]
since \(h\) is increased by \(10 \%, \frac{\delta h}{h}=0.1\)
Question 5
4 / -1

If the distance between a point charge and an infinitely large charge plate is increased by factor 2, the new force on the point charge will be

A
decreased by factor 4

B
increased by factor 2

C
decreased by factor 2

D
the same

Solution

Since an infinitely large charged plate will create an uniform field (constant), therefore the force remains the same.
Question 6
4 / -1

Two point charges, each of charge + q, are fixed at (+ a, 0) and (– a, 0). Another positive point charge q placed at the origin is free to move along the x-axis. The charge q at the origin in equilibrium will have

A
maximum force and minimum potential energy

B
minimum force and maximum potential energy

C
maximum force and maximum potential energy

D
minimum force and minimum potential energy

Solution

The net force on q at the origin is \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}=\frac{1}{4 \pi \xi} \cdot \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \mathrm{i}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}}(-\hat{\mathrm{i}})=\hat{0}\)
The P.E. of the charge \(q\) in between the extreme charges at a distance \(x\) from the
origin along + ve \(\mathrm{x}\) -axis is \(U=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{x}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{\left(a^{2}+x\right)}\)
\(=\frac{1}{4 \pi \varepsilon_{0}} \cdot q^{2}\left[\frac{1}{a-x}+\frac{1}{a^{2}+x}\right]\)
\(\frac{d U}{d x}=\frac{q^{2}}{4 \pi \varepsilon_{0}}\left[\frac{1}{(a-x)^{2}}+\frac{1}{\left(a^{2}+x\right)^{2}}\right]\)
For U to be minimum
\(\frac{d U}{d x}=0, \frac{d^{2} U}{d x^{2}}>0\)
\(\Rightarrow(a-x)^{2}=(a+x)^{2}\)
\(\Rightarrow a+x=\pm(a-x)\)
\(\Rightarrow x=0,\) because other solution is irrelevant.
Thus, the charged particle at the origin will have minimum force and minimum
potential energy.
Question 7
4 / -1

Three rods made of the same material and having the same cross-section have been joined as shown in the figure below. Each rod is of the same length. The left and right ends are kept at 0^oC and 90^oC, respectively. The temperature of the junction of the three rods will be

A
45^oC

B
60^oC

C
30^oC

D
20^oC

Solution

Let \(A\) and \(l\) be the area of cross-section and the length of each rod. If \(k\) is the coefficient of thermal conductivity and \(f^{\circ} \mathrm{C}\) the temperature of the junction \(O,\) then the rates at which heat energy enters \(O\) from rods \(A\) and \(B\) are

\[
Q_{A}=\frac{k A(90-t)}{l}
\]
and \(\quad Q_{B}=\frac{k A(90-t)}{l}\)
The rate at which heat energy flows in rod \(C\) is
\[
Q_{C}=\frac{k A(t-0)}{l}
\]
In the steady state, rate at which heat energy enters \(O=\) rate at which heat energy leaves \(O\), i.e.
\[
Q_{A}+Q_{B}=Q_{C}
\]
or \(\quad \frac{k A(90-t)}{1}+\frac{k A(90-t)}{1}=\frac{k A(t-0)}{l}\)
or \(\quad(90-t)+(90-t)=t\)
or \(\quad 3 t=180\) or \(t=60^{\circ} \mathrm{C}\). Hence the correct choice is
(2).
Question 8
4 / -1

Consider a logic circuit with three inputs A, B and C. The values assigned to A, B and C are such that \(\overline{\mathrm{ABC}}=\) \(\overline{\mathrm{A}+\mathrm{B}+\mathrm{C}} \cdot\) Then \(\overline{\mathrm{A}} \mathrm{BC}+\mathrm{A} \overline{\mathrm{B}} \mathrm{C}+\mathrm{AB} \overline{\mathrm{C}}\) is equal to

A
A + B + C

B
\(\overline{\mathrm{ABC}}\)

C
0

D
1

Solution

For the given condition, i.e. \(\overline{\mathrm{ABC}}=\overrightarrow{\mathrm{A}+\mathrm{B}+\mathrm{C}},\) only two sets of inputs can work, i.e. \(\mathrm{A}=\mathrm{B}=\mathrm{C}=1\) or \(\mathrm{A}=\mathrm{B}=\mathrm{C}=0\)
On substituting \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) in given expression, \(\overline{\mathrm{A}} \mathrm{BC}+\mathrm{A} \overrightarrow{\mathrm{B}} \mathrm{C}+\mathrm{AB} \overline{\mathrm{C}},\) we get result as \(0 .\)
Question 9
4 / -1

The resistance of bulb filament is 100W at a temperature of 100°C. If its temperature coefficient of resistance is 0.005 per °C, its resistance will become 200 W at a temperature (°C) of _____.

A
200

B
100

C
150

D
300

Solution

100 = R₀(1 + 0.005 × 100) -------- (i)

200 = R₀(1 + 0.005 × t) --------- (ii)

Dividing (i) by (ii) and solving

t = 200 °C
Question 10
4 / -1

A liquid takes 6 minutes to cool from 80^oC to 50^oC. If the temperature of the surroundings is 20^oC, then how long will it take to cool from 60^oC to 30^oC?

A
6 minutes

B
8 minutes

C
10 minutes

D
12 minutes

Solution

Given
\[
\log _{e}\left(\frac{80-20}{50-20}\right)=6 K \text { or } \log _{e}(2)=6 K
\]
If the body takes \(t\) minutes to cool from \(60^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C},\) then
\[
\log _{e}\left(\frac{60-20}{30-20}\right)=t K \text { or } \log _{e}(4)=t K
\]
Dividing ( 2) by (1), we have
\[
\frac{t}{6}=\frac{\log _{e}(4)}{\log _{e}(2)}=\frac{2 \log _{e}(2)}{\log _{e}(2)}=2
\]
or \(t=12\) minutes.
Hence the correct choice is (4) .

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Physics Test - 28

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Physics Test - 28

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Question 1

1.08 mA

0.08 mA

1 mA

1 A

Solution

Question 2

700°C

350°C

1400°C

no neutral temperature is possible for this thermocouple

Solution

Question 3

0.99g

0.8g

1.01g

0.9g

Solution

Question 4

20%

15%

10%

5%

Solution

Question 5

decreased by factor 4

increased by factor 2

decreased by factor 2

the same

Solution

Question 6

maximum force and minimum potential energy

minimum force and maximum potential energy

maximum force and maximum potential energy

minimum force and minimum potential energy

Solution

Question 7

45oC

60oC

30oC

20oC

Solution

Question 8

A + B + C

\(\overline{\mathrm{ABC}}\)

0

1

Solution

Question 9

200

100

150

300

Solution

Question 10

6 minutes

8 minutes

10 minutes

12 minutes

Solution

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45^oC

60^oC

30^oC

20^oC