Physics Test

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Physics Test - 14

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Weekly Quiz Competition

Question 1
4 / -1

Using mass (M), length (L), time (T) and electric current (A) as fundamental quantities, the dimensions of permittivity will be

A
MLT-1A-1

B
MLT-2A-2

C
M-1L-3T4A2

D
M2L-2T-2A2

Solution

Permittivity has following dimensions: \([\epsilon]=\frac{\left[q^{2}\right]}{[F]\left[r^{2}\right]}=\frac{\left[C^{2}\right]}{\left[M^{1} L^{1} T^{-2}\right]\left[M^{0} L^{2} T^{0}\right]}\)
\([\epsilon]=\frac{\left[A^{2} T^{2}\right]}{\left[M^{1} L^{3} T^{-2}\right]}=\left[M^{-1} L^{-3} A^{2} T^{4}\right]\)
Question 2
4 / -1

A heater wire consumes power `P` when connected to a 220V source. If the wire is cut into four equal pieces, which are now connected in parallel, find the power consumed by the combination, if connected to the same power supply.

A
4P

B
P/4

C
16P

D
P/16

Solution

Let the resistance of the wire be R:
So power consumed \(p=\frac{y^{2}}{R} \cdots \cdots \cdots(1)\)
When resistance \(R\) is divided in to 4 parts, resistance of each part is \(\frac{R}{4},\) so resulting resistance \(\frac{1}{R^{\prime}}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R}\)
Or \(\mathrm{R}=\mathrm{R} / 16\)
So now power \(=\frac{v^{2}}{R / 16}=16 p\)
So new power is \(16 p ;\) or \(3^{r d}\) option is the answer
Question 3
4 / -1

The apparent depth of water in cylindrical water tank of diameter '2R' cm is reducing at the rate of 'x' cm/min when water is being drained out at a constant rate. The amount of water drained in cc per minute is _____. (n1= refractive index of air and n2= refractive index of water)

A
\(\frac{x \pi R^{2} n_{1}}{n_{2}}\)

B
\(\frac{x \pi R^{2} n_{2}}{n_{1}}\)

C
\(\frac{2 \pi R^{2} n_{1}}{n_{2}}\)

D
πR2x

Solution

\(\frac{\mathrm{h}}{\mathrm{h}^{\prime}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}},\) where \(\mathrm{h}\) is the real depth and \(\mathrm{h}^{\prime}\) is the apparent depth.
\(\mathrm{h}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \times \mathrm{h}^{\prime}\)
Differentiating w.r.t. time
\(\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}} \times \frac{\mathrm{dh}^{\prime}}{\mathrm{dt}}\)
Amount of water drained \(=\) Area \(\times \frac{\mathrm{dh}^{\prime}}{\mathrm{dt}}=\pi \mathrm{R}^{2} \mathrm{x}\)
Therefore, \(\frac{d h}{d t}=\frac{n_{2}}{n_{1}}\left(\pi R^{2} x\right)\)
Question 4
4 / -1

Directions: In the following question, a statement of Assertion is given, followed by a corresponding statement of Reason. Mark the correct answer as:

(a) If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

(b) If both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

(c) If Assertion is true, but Reason is false.

(d) If both Assertion and Reason are false.

Assertion : Neutrons penetrate a matter more readily as compared to protons.

Reason : Neutrons are slightly more massive than protons.

A
A

B
B

C
C

D
D

Solution

Here assertion given is correct. The reason given for the assertion is correct but it is not the reason for the assertion.
Question 5
4 / -1

Consider the followingtwo statements A and B and identify the correct answer.

A:Constantan-Copper thermo-couple is generally used to measure temperature upto 1600oC.

B:In an iron-copper thermocouple, current flows from iron to copper through cold junction.

A
Both A and B are true.

B
Both A and B are false.

C
A is true but B is false.

D
A is false but B is true.

Solution

Both A and B are true. In an iron-copper thermocouple, current flows from iron to copper since iron comes first in see back list.
Question 6
4 / -1

A can is taken out from a refrigerator at \(0^{\circ} \mathrm{C}\). The atmospheric temperature is \(25^{\circ} \mathrm{C}\). If \(t_{1}\) is the time taken to heat from \(0^{\circ} \mathrm{C}\) to \(5^{\circ}\) and \(t_{2}\) is the time taken from \(10^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C},\) then

A
t1> t2

B
t1< t

C
t₁ = t₂

D
there is no relation

Solution

Sir Newton's law of heating states that rate of change of temperature of a object is proportional to the difference in temperatures of the surrounding and the object. In the given cases, when initial temperature of can is \(0^{\circ} \mathrm{C}\), the difference in temperatures of can

and surrounding is \(25-0=25^{\circ} \mathrm{C}\).

When initial temperature of can is \(10^{\circ} \mathrm{C}\), the difference in temperatures of can and surrounding

is \(25-10=15^{\circ} \mathrm{C}\) therefore the rate of heating will be higher, when temperature difference is

\(25^{\circ} \mathrm{C}\) i.e. time \(t_{1}\) will be smaller than \(\mathrm{t}_{2}\)

\(t_{1}
Question 7
4 / -1

A beam of protons with velocity 4 x 10⁵ m/s enters a uniform magnetic field of 0.3 T at an angle of 60o to the magnetic field. Find the radius of the helical path taken by the proton beam.

A
0.2 cm

B
1.2 cm

C
2.2 cm

D
0.122 cm

Solution

When the charged particle is moving at an angle to the field (other than \(0^{\circ}, 90^{\circ}\) and \(180^{\circ}\) ), in this situation resolving the velocity of the particle along and perpendicular to the field, we find that the particle moves with constant velocity \(v \cos \theta\) along the field and at the same time it is also moving with velocity \(v\) sin \(\theta\) perpendicular to the the field due to which it will describe a circle (in a plane perpendicular to the field) of radius
\[
r=\frac{m(v \sin \theta)}{q B}
\]
Here, \(\quad m=1.67 \times 10^{-27} \mathrm{kg}\)
\[
\begin{array}{l}
\qquad \begin{aligned}
v &=4 \times 10^{5} \mathrm{m} / \mathrm{s} \\
\theta &=60^{\circ} \\
q &=1.6 \times 10^{-19} \mathrm{C} \\
B &=0.3 \mathrm{T}
\end{aligned} \\
\therefore \quad r=\frac{1.67 \times 10^{-27} \times 4 \times 10^{5} \times(\sqrt{3} / 2)}{1.6 \times 10^{-19} \times 0.3}
\end{array}
\]
\(=0.012 \mathrm{m}\)
\[
=1.2 \mathrm{cm}
\]
Question 8
4 / -1

If power dissipated by 5Ω resistor is 20 W, then find the power dissipated by 4Ω resistor.

A
10 W

B
20 W

C
4 W

D
15 W

Solution

since the resistances \(4 \Omega, 6 \Omega\) and \(5 \Omega\) are in parallel, therefore the voltage across them will be the
same. We know, \(P=\frac{V^{2}}{R}\)
\(\mathrm{P} \times \mathrm{R}=\mathrm{V}^{2}\)
Across the \(5 \Omega\) resistance,
\(20 \times 5=100=v^{2}\)
\(\mathrm{V}=10 \mathrm{Volts}\)
Current passing through the upper arm (whose equivalent resistance is \(10 \Omega)=\frac{10}{10}=1 \mathrm{A}(\cup \operatorname{sing} \mathrm{V}=\) \(|R\rangle\)
So, power dissipated by \(4 \Omega\) resistor \(=1^{2} R=4 \mathrm{W}\)
Question 9
4 / -1

The resultant of forces \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) is \(\overrightarrow{\mathrm{R}}\). If \(\overline{\mathrm{Q}}\) is doubled, then \(\overrightarrow{\mathrm{R}}\) is doubled. If the direction of \(\overline{\mathrm{Q}}\) is reversed, then \(\overrightarrow{\mathrm{R}}\) is again doubled.

then \(\mathrm{P}^{2}: \mathrm{Q}^{2}: \mathrm{R}^{2}\) is

A
3 : 1 : 1

B
2 : 3 : 2

C
1 : 2 : 3

D
2 : 3 : 1

Solution

\(R^{2}=P^{2}+Q^{2}+2 P Q \cos \theta\)
\(4 R^{2}=P^{2}+4 Q^{2}+4 P Q \cos \theta\)
\(4 R^{2}=P^{2}+Q^{2}-2 P Q \cos \theta\)
\(\mathrm{On}(\mathrm{i})+(\mathrm{ii}), 5 R^{2}=2 P^{2}+2 Q^{2}\)
On (iii) \(\times 2+(\) ii \(), 12 R^{2}=3 P^{2}+6 Q^{2}\)
\(2 P^{2}+2 Q^{2}-5 R^{2}\)
\[
3 P^{2}+6 Q^{2}-12 R^{2}
\]
by cross multiplication
\[
\frac{P^{2}}{-24+30}=\frac{Q^{2}}{24-15}=\frac{R^{2}}{12-6}
\]
Question 10
4 / -1

Two men carry a weight of 240 kg between them by means of two ropes fixed to the weight. One rope is inclined at 60× to the vertical and the other at 30×. The tensions in the ropes are

A
120 kg, 120 kg

B
120√3 kg, 120 kg

C
120√3 kg, 120√3 kg

D
None of these

Solution

\(\mathrm{mg}=\mathrm{T}_{2} \cos 60^{\circ}+\mathrm{T}_{1} \cos 30^{\circ} \ldots \ldots\)
\(\mathrm{T}_{1} \sin 30^{\circ}=\mathrm{T}_{2} \sin 60^{\circ} \ldots \ldots\) (ii)
\(\Rightarrow T_{1} / T_{2}=\sqrt{3}\)
\(\mathrm{mg}=\mathrm{T}_{2} \cos 60^{\circ}+\mathrm{T}_{1} \cos 30^{\circ}\)
\(\frac{240}{T_{2}}=\frac{1}{2}+\frac{T_{1}}{T_{2}} \times \frac{\sqrt{3}}{2}\)
Using eq (iii) we get, \(T_{2}=120 \mathrm{kg}\) and \(T_{1}=120 \sqrt{3} \mathrm{kg}\)

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Physics Test - 14

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Physics Test - 14

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Question 1

MLT-1A-1

MLT-2A-2

M-1L-3T4A2

M2L-2T-2A2

Solution

Question 2

4P

P/4

16P

P/16

Solution

Question 3

\(\frac{x \pi R^{2} n_{1}}{n_{2}}\)

\(\frac{x \pi R^{2} n_{2}}{n_{1}}\)

\(\frac{2 \pi R^{2} n_{1}}{n_{2}}\)

πR2x

Solution

Question 4

A

B

C

D

Solution

Question 5

Both A and B are true.

Both A and B are false.

A is true but B is false.

A is false but B is true.

Solution

Question 6

t1> t2

t1< t

t1 = t2

there is no relation

Solution

Question 7

0.2 cm

1.2 cm

2.2 cm

0.122 cm

Solution

Question 8

10 W

20 W

4 W

15 W

Solution

Question 9

3 : 1 : 1

2 : 3 : 2

1 : 2 : 3

2 : 3 : 1

Solution

Question 10

120 kg, 120 kg

120√3 kg, 120 kg

120√3 kg, 120√3 kg

None of these

Solution

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t₁ = t₂