Physics Test

Result

Physics Test - 36

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A wire of length \(1.0 \mathrm{m}\) and radius \(10^{-3} \mathrm{m}\) is carrying a heavy current and is assumed to radiate as a black body. At equilibrium, its temperature is \(900 \mathrm{K}\) while that of the surroundings is \(300 \mathrm{K}\). The resistivity of the material of the wire at \(300 \mathrm{K}\) is \(\pi^{2} \times 10^{-8} \mathrm{ohm}-\mathrm{m}\) and its temperature coefficient of resistance is \(7.8 \times \frac{10^{-3}}{c^{\circ}}\). Find the current in the wire. [Given Stefan's constant \(=5.68 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4}\) ]

A
18 A

B
36 A

C
54 A

D
72 A

Solution
Question 2
1 / -0

The fundamental frequency of vibration of 1 cm long string is 256 Hz. If the length of the string is reduced to 1/4cm keeping the tension unaltered, the new fundamental frequency will be :

A
256 Hz

B
512 Hz

C
1024 Hz

D
2048 Hz

Solution

We know that, \(n \propto \frac{1}{l}\) \(\therefore \frac{n_{2}}{n_{1}}=\frac{l_{1}}{l_{2}}\)
\(n_{2}=\frac{l_{1}}{l_{2}} \times n_{1}=\frac{1 \times 256}{1 / 4}=1024 H z\)
Hence, the correct option is (C)
Question 3
1 / -0

A string of density 7.5g/cm⁻³ and area of cross-section 0.2mm² is stretched under a tension of 20N. When it is plucked at the mid point, the speed of the transverse waves on the wire is:

A
116 m/s

B
40 m/s

C
200 m/s

D
0 m/s

Solution

Density of string \(\rho=7.5 \mathrm{g} / \mathrm{cm}^{3}\)
Cross-sectional area \(A=0.2 \mathrm{mm}^{2}=0.002 \mathrm{cm}^{2}\)
Thus mass per unit length of string \(\mu=\rho A=(7.5)(0.002)=0.015 g / \mathrm{cm}\)
\(\Longrightarrow \mu=0.015 \mathrm{g} / \mathrm{cm}=0.0015 \mathrm{kg} / \mathrm{m}\)
Tension in the string \(T=20 N\)
Speed of transverse wave \(v=\sqrt{\frac{T}{\mu}}\)
\(\therefore v=\sqrt{\frac{20}{0.0015}}=116 \mathrm{m} / \mathrm{s}\)
Hence, the correct option is (A)
Question 4
1 / -0

A sound signal is sent through a composite tube as shown in the Fig. The radius of the semicircle is r. The speed of sound in air is v. The source of sound is capable of generating frequencies in the range f₁ to f₂ (f₂ > f₁). If nn is an integer, frequency for maximum intensity is given by :

A
nv/r

B
nv/r(π − 2)

C
nv/πr

D
nv/(r − 2)π

Solution

Path difference
\(\pi r-2 r=n \lambda\) or
\(r(\pi-2)=\frac{n v}{f}\)
thus \(f=\frac{n v}{r(\pi-2)}\)
Hence, the correct option is (B)
Question 5
1 / -0

When a stationary wave is formed, then its frequency is :

A
Same as that of the individual waves

B
Twice that of the individual waves

C
Half that of the individual waves

D
That of the individual waves

Solution

Let us consider a progressive wave of amplitude \(a\), wavelength \(\lambda\) and frequency \(f=\frac{1}{T}\) travelling in the direction of \(X\) -axis whose equation is given by\(y_{1}=a \sin \left(2 \pi\left[\frac{t}{T}-\frac{x}{\lambda}\right]\right)\)
This wave is reflected from a free end and it travels in the negative direction of \(X\) axis with frequency \(f=\frac{1}{T}\) \(y_{1}=a \sin \left(2 \pi\left[\frac{t}{T}+\frac{x}{\lambda}\right]\right)\)
According to principle of superposition \(\Rightarrow y_{1}+y_{2}=a \sin \left(2 \pi\left[\frac{t}{T}-\frac{x}{\lambda}\right]\right)+a \sin \left(2 \pi\left[\frac{t}{T}+\frac{x}{\lambda}\right]\right)\)
\(\Rightarrow y_{1}+y_{2}=a \sin \left(\frac{2 \pi t}{T}\right) \cos \left(\frac{2 \pi x}{\lambda}\right)\)
Here we can see the frequency of resultant wave is also \(f=\frac{1}{T}\)
Hence, the correct option is (A)
Question 6
1 / -0

A uniform string is vibrating with a fundamental frequency 'f'. The new frequency, if radius and length both are doubted would be:

A
2 f

B
3 f

C
f/4

D
f/3

Solution

Wavelength, \(\lambda=2 L\) Speed of wave, \(v=\sqrt{\frac{T L}{m}}\) \(v=f \lambda\)
\(\Longrightarrow f \propto \frac{1}{\sqrt{m L}}\)
\(m_{2}=8 m_{1}\) since volume is increased to 8 times
\(L_{2}=2 L_{1}\)
\(\Longrightarrow f_{2}=f / 4\)
Hence, the correct option is (C)
Question 7
1 / -0

A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per seconds, when sounded with a source of frequency 513 Hz. The unknown frequency is :

A
254 Hz

B
246 Hz

C
240 Hz

D
260 Hz

Solution

Let the unknown frequency be \(f\). Beat frequency \(b=\left|f-f^{\prime}\right|\) For \(f^{\prime}=250 \mathrm{Hz} \quad b=4\)
\(\therefore|f-250|=4\)
\(\ldots\)
For \(f^{\prime}=513 \quad b=5\)
\(\therefore|2 f-513|=5\)
Case \(1: 2 f>513 H z\) and \(f>250 H z\)
We get \(f-250=4 \Rightarrow f=254 H z\)
and \(2 f-513=5 \quad \Longrightarrow f=259 H z\)
which is not possible. Case \(2: 2 f<513 H z\) and \(f>250 H z\)
We get \(f-250=4 \quad \Longrightarrow \quad f=254 H z\)
And \(513-2 f=5 \quad \Longrightarrow \quad f=254 H z\)
Case \(3: 2 f<513 H z\) but \(f<250 H z\)
We get \(250-f=4 \quad \Longrightarrow \quad f=246 H z\)
And \(513-2 f=5 \quad \Longrightarrow \quad f=254 H z\)
Which is not possible. So the unkown frequency \(f=254 \mathrm{Hz}\)
Hence, the correct option is (A)
Question 8
1 / -0

A car is moving at a velocity of 17 ms⁻¹ towards an approaching bus that blows a horn at a frequency of 640 Hz on a straight track. The frequency of this horn appears to be 680 Hz to the car driver. If the velocity of sound in air is 340 ms⁻¹, then the velocity of the approaching bus is:

A
2 ms⁻¹

B
4 ms⁻¹

C
8 ms⁻¹

D
10 ms⁻¹

Solution

Given velocity of sound \(v=340 \mathrm{m} / \mathrm{s}\)
Velocity of listner \(v_{L}=17 \mathrm{m} / \mathrm{s}\) Velocity of source \(=v_{S}\) Frequency of horn emitted \(\nu=640 \mathrm{Hz}\)
The apparent frequency \(\nu^{\prime}=\nu \frac{\left(v+v_{L}\right)}{v-v_{S}}\)
\(680=640\left(\frac{340+17}{340-v_{s}}\right)\)
On solving we get \(v_{S}=4 \mathrm{m} / \mathrm{s}\)
Hence, the correct option is (B)
Question 9
1 / -0

The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe is 100 Hz. The frequency of the third harmonic of the closed pipe is:

A
100 H z

B
200 H z

C
300 H z

D
150H z

Solution

Let the length of closed pipe be
\(L^{\prime}\)
Thus length of open pipe \(L=2 L^{\prime}\)
Fundamental frequency in open pipe \(f_{o}=\frac{v}{2 L}=\frac{v}{4 L^{\prime}}=100 \mathrm{Hz}\)
Frequency of third harmonic in closed pipe \(f_{3}=\frac{3 v}{4 L^{\prime}}\)
\(\Longrightarrow f_{3}=3 \times 100=300 \mathrm{Hz}\)
Hence, the correct option is (C)
Question 10
1 / -0

A block of wood of side 40 cm floats in water in such a way that it lower face is 5 cm below the free surface of water. What is the weight of the blocks?

A
64 kg

B
16 kg

C
8 kg

D
Cannot be determined as density of wood is not given

Solution

Volume of part of wooden block immersed in water \(V_{\text {in}}=(40 \times 40 \times 5) \mathrm{cm}^{2}=(40 \times 40 \times 5) \times 10^{-6} \mathrm{m}^{3}\)
Density of water \(\sigma_{w}=10^{3} \mathrm{kg} / \mathrm{m}^{3}\)
Weight of wooden block \(=\) Weight of water displaced \(M g=V_{i n} \sigma_{w} g\)
\(M g=(40 \times 40 \times 5) \times 10^{-6} \times 10^{3} g\)
\(\Rightarrow M=8 \mathrm{kg}\)
Hence, the correct option is (C)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 36

Result

Physics Test - 36

Score

-

Rank

-

SHARING IS CARING

Question 1

18 A

36 A

54 A

72 A

Solution

Question 2

256 Hz

512 Hz

1024 Hz

2048 Hz

Solution

Question 3

116 m/s

40 m/s

200 m/s

0 m/s

Solution

Question 4

nv/r

nv/r(π − 2)

nv/πr

nv/(r − 2)π

Solution

Question 5

Same as that of the individual waves

Twice that of the individual waves

Half that of the individual waves

That of the individual waves

Solution

Question 6

2 f

3 f

f/4

f/3

Solution

Question 7

254 Hz

246 Hz

240 Hz

260 Hz

Solution

Question 8

2 ms−1

4 ms−1

8 ms−1

10 ms−1

Solution

Question 9

100 H z

200 H z

300 H z

150H z

Solution

Question 10

64 kg

16 kg

8 kg

Cannot be determined as density of wood is not given

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.

2 ms⁻¹

4 ms⁻¹

8 ms⁻¹

10 ms⁻¹