Physics Test

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Physics Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A disc is given an initial angular velocity ω0ω0 and placed on rough horizontal surface as shown. The quantities which will not depend on the coefficient of friction is/are

A
The time until rolling begins

B
The displacement of the disc until rolling begins

C
The velocity when rolling begins

D
The work done by the force of friction

Solution

velocity of the disc when rolling begins can be obtained using the conservation of angular momentum principle about the point through which the friction force acts. So, the coefficient of friction has no bearing on final velocity. The work done by the force of friction will simply be change in kinetic energy.

∴ option 3 is correct.
Question 2
1 / -0

A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is

A
$\frac{M ω^{2} L}{2}$

B
$M L ω^{2}$

C
$\frac{M L ω^{2}}{4}$

D
$\frac{M L^{2} ω 2}{2}$

Solution

Taking a small element : thus the force excreted by this element is dF then.

$\mathrm{dF}=(\mathrm{dm}) \omega^{2} \mathrm{x}$
$\mathrm{dm}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{dx}$

integrating we get

$\int_{0}^{\mathrm{F}} \mathrm{d} \mathrm{F}=\int_{0}^{\mathrm{L}} \frac{\mathrm{M}}{\mathrm{L}} \omega^{2} \mathrm{x} \mathrm{d} \mathrm{x}$
$F=\frac{M}{L} \omega^{2}\left(\frac{x^{2}}{2}\right)_{0}^{L}$
$\mathrm{F}=\frac{\mathrm{M} \omega^{2} \mathrm{L}}{2}$
Question 3
1 / -0

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R.B.=2400 Ω. R.B.=2400 Ω. Deflection becomes 20 divisions when resistance taken from resistance box is 4900 Ω. Then we can conclude :

Note: This question is awarded as the bonus. Now the question is corrected.

A
Current sensitivity of galvanometer is 20 μA/division

B
Resistance of galvanometer is 200 Ω.

C
Resistance required on R.B. for a deflection of 10 divisions is 9800 Ω.

D
Full scale deflection current is 2 mA.

Solution

Current in the circuit,

$I = \frac{V}{2400 + R^{G}} a n d \frac{I}{2} = \frac{V}{4900 + R G}$

⇒ R_G= 100Ω

$I = \frac{2}{2500} = 8 \times 10^{- 4} A$

Current Sensitivity = $\frac{8 \times 10^{- 4}}{40}$

20μA per division
Question 4
1 / -0

A particle is released from rest from a tower of height 3h. The ratio of time taken to fall equal heights h i.e., t₁:t₂:t₃is

A
$\sqrt{3} : \sqrt{2} : 1$

B
$3 : 2 : 1$

C
$9 : 4 : 1$

D
$1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$

Solution

$h = \frac{1}{2} g t_{1}^{2}$

$2 h = \frac{1}{2} g {(t_{1} + t_{2})}^{2}$

and $3 h = \frac{1}{2} g {(t_{1} + t_{2} + t_{3})}^{2}$

i.e.,t₁: (t₁+ t₂) : (t₁+ t₂+ t₃) = 1 : √2 : √3

or t₁: t₂: t₃= 1 : (√2−1) : (√3−√2)
Question 5
1 / -0

What is the spring constant for the combination of springs shown in figure?

A
k

B
2k

C
4k

D
$\frac{5 k}{2}$

Solution

k_eq of the system k_eq= 2k+k+k

(Since all the springs are effectively in parallel)

k_eq = 4k
Question 6
1 / -0

A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the center of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. What is the energy per photon in (eV) associated with the sodium light?

A
2.11

B
2.22

C
1

D
2

Solution

Given, power of lamp, P = 100 W

Wavelength of the sodium light, λ=589 nm = 589 × 10⁻⁹ m

Planck constant h = 6.63 × 10^-34 J-s

Energy of each photon

$E = \frac{h c}{λ} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{589 \times 10^{- 9}}$

(∵ c=3×10⁸ m/s)

=3.38×10⁻¹⁹ J

= $\frac{3.38 \times 10^{- 19}}{1.6 \times 10^{- 19}} e V$

= 2.11 eV
Question 7
1 / -0

Two long parallel conductors are placed at right angles to a metre scale at the 2 cm and 6 cm marks as shown in the figure

They carry currents of 1 A and 3 A respectively. They will produce zero magnetic field at (ignore earth's magnetic field)

A
5 cm mark

B
3 cm mark

C
1 cm mark

D
8 cm mark

Solution

Here, x is the distance from the 2 cm mark.

$\frac{\mu_{o} I_{1}}{2 \pi x}=\frac{\mu_{o} I_{2}}{2 \pi(4-x)}$
$(4-x) \times 1=x \times 3$
$4-x=3 x$
$4=4 x \Rightarrow x=1$
Question 8
1 / -0

If the nucleus ${}_{13}^{27}A l$ has a nuclear radius 3.6 fm, then ${}_{32}^{125}T e$ would have its radius approximately as

A
9.6 fm

B
12.0 fm

C
4.8 fm

D
6.0 fm

Solution

Nuclear radius, R=(R₀)A^1/3

where A is the mass number.

$\therefore \quad \frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=\left(\frac{A_{\mathrm{Te}}}{A_{\mathrm{Al}}}\right)^{1 / 3}=\left(\frac{125}{27}\right)^{1 / 3}=\left(\frac{5}{3}\right)$
or $\quad R_{\mathrm{Te}}=\frac{5}{3} \times R_{\mathrm{Al}}=\frac{5}{3} \times 3 \cdot 6=6 \mathrm{fm}$

(Given R_Al=3⋅6 fm)
Question 9
1 / -0

A refrigerator is to maintain the eatables, kept inside it, at 9^oC. The coefficient of performance of the refrigerator, if the room temperature is 36^oC, is

A
10.45

B
4

C
3

D
11.45

Solution

Here, T₁=36℃=36+273=309 K

T₂=9℃=9+273=282 K

The coefficient of performance of a refrigerator is given by

cop= $\frac{T_{2}}{T_{1} - T_{2}} = \frac{282}{309 - 282} = \frac{282}{27} = 10.45$
Question 10
1 / -0

A point mass mm is suspended at the end of a massless wire of length L and cross-section area A. If Y is Young's modulus for the wire, then the frequency of oscillations for the SHM along the vertical line is

A
$\frac{1}{2 π} \sqrt{\frac{Y A}{m L}}$

B
$2 π \sqrt{\frac{m L}{Y A}}$

C
$\frac{1}{π} \sqrt{\frac{Y A}{m L}}$

D
$π \sqrt{\frac{m L}{Y A}}$

Solution

In this case,

Stress = $\frac{m g}{A}$

Strain = l/L (where l is extension)

Now, Young's modulus Y is given by

Y= $\frac{s t r e s s}{s t r a i n} = \frac{m g / A}{l / L}$

mg = $\frac{Y A l}{L}$

So, kl=YAl/L (∵ mg=kl)

(k is force constant)

Now, frequency is given by

$n = \frac{1}{2 π} \sqrt{\frac{k}{m}}$

= $\frac{1}{2 π} \sqrt{(\frac{Y A}{m L})}$

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Physics Test - 19

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Physics Test - 19

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SHARING IS CARING

Question 1

The time until rolling begins

The displacement of the disc until rolling begins

The velocity when rolling begins

The work done by the force of friction

Solution

Question 2

Mω2L2

MLω2

MLω24

ML2ω22

Solution

Question 3

Current sensitivity of galvanometer is 20 μA/division

Resistance of galvanometer is 200 Ω.

Resistance required on R.B. for a deflection of 10 divisions is 9800 Ω.

Full scale deflection current is 2 mA.

Solution

Question 4

3:2:1

3:2:1

9:4:1

1:2-1:3-2

Solution

Question 5

k

2k

4k

5k2

Solution

Question 6

2.11

2.22

1

2

Solution

Question 7

5 cm mark

3 cm mark

1 cm mark

8 cm mark

Solution

Question 8

9.6 fm

12.0 fm

4.8 fm

6.0 fm

Solution

Question 9

10.45

4

3

11.45

Solution

Question 10

12πYAmL

2πmLYA

1πYAmL

πmLYA

Solution

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$\frac{M ω^{2} L}{2}$

$M L ω^{2}$

$\frac{M L ω^{2}}{4}$

$\frac{M L^{2} ω 2}{2}$

$\sqrt{3} : \sqrt{2} : 1$

$3 : 2 : 1$

$9 : 4 : 1$

$1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$

$\frac{5 k}{2}$

$\frac{1}{2 π} \sqrt{\frac{Y A}{m L}}$

$2 π \sqrt{\frac{m L}{Y A}}$

$\frac{1}{π} \sqrt{\frac{Y A}{m L}}$

$π \sqrt{\frac{m L}{Y A}}$