Physics Test

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Physics Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

In the given circuit the average power developed is -

A
$50 \sqrt{2} w a t t$

B
$200 w a t t$

C
$150 \sqrt{2} w a t t$

D
$200 \sqrt{2} w a t t$

Solution

P = V_rms I_rms cosϕ

P = $V_{r m s} \frac{V_{r m s}}{Z} \frac{R}{Z}$

= $\frac{{V^{2}}_{r m s}}{Z_{2}} R$

Z= $\sqrt{R^{2} + {(L ω)}^{2}} = \sqrt{{(50)}^{2} + {(0.2 \times 250)}^{2}}$

= $\sqrt{2500 + {(50)}^{2}} = 50 \sqrt{2}$

∴ P = ${(\frac{200}{\sqrt{2}})}^{2} \times \frac{50}{50 \sqrt{2}} \times \frac{1}{50 \sqrt{2}}$

=200 watt
Question 2
1 / -0

Two balls with equal charges are in a vessel with ice at −10 °C at a distance of 25 cm from each other. On forming water at 0 °C, the balls are brought nearer to 5 cm for the interaction between them to be same. If the dielectric constant of water at 0 °C is 80, the dielectric constant of ice at −10 °C is

A
40

B
3.2

C
20

D
6.4

Solution

$(- 10 ° C i c e) (0 ° C W a t e r)$
$q 1 \frac{k_{1}}{r_{1} = 25 c m} q 2 q 1 \frac{k_{2} = 80}{r_{2} = 5 c m} q 2$
Given F_ice= F_water
$\frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{k_{1} r_{1}^{2}} = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{k_{2} r_{1}^{2}}$
$\therefore \mathrm{k}_{1} \mathrm{r}_{1}^{2}=\mathrm{k}_{2} \mathrm{r}_{2}^{2}$
$\therefore \mathrm{k}_{1}=\mathrm{k}_{2}\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}$
$=80 \times\left(\frac{5}{25}\right)^{2}=3.2$
Question 3
1 / -0

The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is :

A
69 mA

B
0.69 mA

C
6.9 mA

D
9.6 mA

Solution

$β = 69 = \frac{I C}{I B}$
$α = \frac{β}{1 + β} = \frac{69}{70} = \frac{I C}{I E}$
$I_{C} = I_{E} \times \frac{69}{70}$
= $\frac{69}{70} \times 7$
= 6.9 mA
Question 4
1 / -0

Two electric lamps A and B radiate the same power. Their filaments have the same dimensions and have emissivities e_A and e_B respectively. Their surface temperatures are T_A and T_B. The ratio T_A/T_B will be equal to

A
${(\frac{e_{B}}{e_{A}})}^{1 / 4}$

B
${(\frac{e_{B}}{e_{A}})}^{1 / 2}$

C
${(\frac{e_{A}}{e_{B}})}^{1 / 2}$

D
${(\frac{e_{A}}{e_{B}})}^{1 / 4}$

Solution

The power radiated by a filament is P = eA (σT₄) (where e = emissivity, σ= Stefan's constant and T= surface temperature)
Here, eT⁴= constant or $e_{A} T_{A}^{4} = e_{B} T_{B}^{4}$
Question 5
1 / -0

A car is moving with 90 km/h blows a horn of 150 Hz, towards a cliff. The frequency of the reflected sound heard by the driver will be (speed of sound in air is 340 m/s)

A
150 Hz

B
140 Hz

C
180 Hz

D
174 Hz

Solution

Given, $v_{s} = 90 k m / h = \frac{90 \times 5}{18} m / s = 25 m / s$
υ_s =150 Hz, speed of sound, v=340 m/s
Here, source is observer of reflected sound and moving towards the reflector (the cliff)
Thus, v_o=25 m/s
Since, Doppler's equation for sound,
$v = \frac{v + v_{s}}{v - v_{o}} υ_{s} = \frac{340 + 25}{340 - 25} \times 150$
= $\frac{365}{315} \times 150 = 173.88 H z$
∴ v =174 Hz
Question 6
1 / -0

In the figure shown if the object 'O' moves towards the plane mirror, then the image I (which is formed after successive reflections from M₁ & M₂ respectively) will move:

A
towards right

B
towards left

C
with zero velocity

D
cannot be determined

Solution

Image (O') of object in plane mirror will act as object for concave mirror.

${(V_{I / M})}_{| |} = M_{T}^{2} . {(V_{o / M})}_{| |}$

so image I in concave mirror will move towards right
Question 7
1 / -0

Which of the following statements is false?

A
Kirchhoff's second law represents energy conservation

B
Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude

C
In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed

D
A rheostat can be used as a potential divider

Solution

Null point does not change.
Question 8
1 / -0

When photons of wavelength λ₁ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ₂ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength λ₃ is used then find the stopping potential for this case:

A
$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{λ_{2}} - \frac{1}{λ_{1}}]$

B
$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{2 λ_{2}} - \frac{1}{λ_{1}}]$

C
$\frac{h c}{e} [\frac{1}{λ_{3}} - \frac{1}{λ_{2}} - \frac{1}{λ_{1}}]$

D
$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{2 λ_{2}} - \frac{3}{2 λ_{1}}]$

Solution

$\frac{\mathrm{hc}}{\lambda_{1}}=\frac{\mathrm{hc}}{\lambda_{0}}+\mathrm{eV} \ldots$ (i)
$\frac{\mathrm{hc}}{\lambda_{2}}=\frac{\mathrm{hc}}{\lambda_{0}}+3 \mathrm{eV} \ldots$ (ii)
$\frac{\mathrm{hc}}{\lambda_{3}}=\frac{\mathrm{hc}}{\lambda_{0}}+\mathrm{eV}^{\prime} \ldots$ (iii)

Equation (i) and (ii)

$\frac{3}{2 \lambda_{1}}-\frac{1}{2 \lambda_{2}}=\frac{1}{\lambda_{0}}$
$\frac{\mathrm{hc}}{\lambda_{3}}-\mathrm{hc}\left[\frac{3}{2 \lambda_{1}}-\frac{1}{2 \lambda_{2}}\right]=\mathrm{eV}$
$\frac{\mathrm{hc}}{\mathrm{e}}\left[\frac{1}{\lambda_{3}}-\frac{3}{2 \lambda_{1}}+\frac{1}{2 \lambda_{2}}\right]=\mathrm{V}^{\prime}$
Question 9
1 / -0

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 ms⁻¹to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

A
30 m s⁻¹

B
20 m s⁻¹

C
10 m s⁻¹

D
5 m s⁻¹

Solution

Hence, m₁=10 kg, m₂=4 kg
v₁=14 m s⁻¹, v₂=0
$v_{C M} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{C M} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = 10 m s^{- 1}$
Question 10
1 / -0

A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C (a condition called auto-collimation). If the mirror is now filled with water, the image will be:

A
real, and will remain at C

B
real, and located at a point between C and ∞

C
virtual, and located at a point between C and O

D
real, and located at a point between C and O

Solution

Due so refraction at water surface object will appear beyond C. This image will acts as object for reflection by mirror. So image will be real, and located between C and O

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Physics Test - 16

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Physics Test - 16

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Question 1

502watt

200watt

1502watt

2002watt

Solution

Question 2

40

3.2

20

6.4

Solution

Question 3

69 mA

0.69 mA

6.9 mA

9.6 mA

Solution

Question 4

eBeA1/4

eBeA1/2

eAeB1/2

eAeB1/4

Solution

Question 5

150 Hz

140 Hz

180 Hz

174 Hz

Solution

Question 6

towards right

towards left

with zero velocity

cannot be determined

Solution

Question 7

Kirchhoff's second law represents energy conservation

Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude

In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed

A rheostat can be used as a potential divider

Solution

Question 8

hce1λ3+1λ2-1λ1

hce1λ3+12λ2-1λ1

hce1λ3-1λ2-1λ1

hce1λ3+12λ2-32λ1

Solution

Question 9

30 m s−1

20 m s−1

10 m s−1

5 m s−1

Solution

Question 10

real, and will remain at C

real, and located at a point between C and ∞

virtual, and located at a point between C and O

real, and located at a point between C and O

Solution

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$50 \sqrt{2} w a t t$

$200 w a t t$

$150 \sqrt{2} w a t t$

$200 \sqrt{2} w a t t$

${(\frac{e_{B}}{e_{A}})}^{1 / 4}$

${(\frac{e_{B}}{e_{A}})}^{1 / 2}$

${(\frac{e_{A}}{e_{B}})}^{1 / 2}$

${(\frac{e_{A}}{e_{B}})}^{1 / 4}$

$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{λ_{2}} - \frac{1}{λ_{1}}]$

$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{2 λ_{2}} - \frac{1}{λ_{1}}]$

$\frac{h c}{e} [\frac{1}{λ_{3}} - \frac{1}{λ_{2}} - \frac{1}{λ_{1}}]$

$\frac{h c}{e} [\frac{1}{λ_{3}} + \frac{1}{2 λ_{2}} - \frac{3}{2 λ_{1}}]$

30 m s⁻¹

20 m s⁻¹

10 m s⁻¹

5 m s⁻¹