Physics Test

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Physics Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

Two charges of +1 μC & + 5 μC are placed 4 cm apart, the ratio of the force exerted by both charges on each other will be :

A
1 : 1

B
1 : 5

C
5 : 1

D
25 : 1

Solution

Coulomb's law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
F=kQ1Q2r2
where Q1 and Q2 are the magnitudes of the two charges respectively and r is the distance between them. The proportionality constant k is called the electrostatic constant.
Each charge experiences a force with the same magnitude and so the ratio of the force exerted by both charges on each other will be 1:1.
Hence, the correct option is (A)
Question 2
1 / -0

Two filaments of same length are connected first in series then in parallel. For the same amount of main current flowing, the ratio of the heat produced is:

A
1 : 2

B
4 : 1

C
1 : 4

D
2 : 1

Solution

Let main current be I \& let the resistance of each wire be R.
In the first case heat produced \(=I 2 R t+I 2 R t=2 I 2 R t\)
In the second case heat produced \(=( I 2) 2 R t \times 2= I _{2} Rt _{2}\)
Ratio \(=2 I _{2} RtI _{2} RT _{2}=4: 1\)
Hence, the correct option is (B)
Question 3
1 / -0

Calculate the power dissipated in a \(10 k \Omega\) resistor with a \(5 mA\) current through the resistor.

A
200 mW

B
250 mW

C
100 mW

D
150 mW

Solution

Using \(p=i^{2} \times R\)
\(p=\left(5 \times 10^{-3}\right)^{2} \times 10 \times 10^{3}\)
\(p=250 mW\)
i.e. The power dissipated in the resistor is \(250 mW\).
Hence, the correct option is (B)
Question 4
1 / -0

An electric bulb is designed to draw power P0 at voltage V0. If the voltage is V, it draws a power P, then-

A
P=(V/V0)P0

B
P=(V0/V)P0

C
P=(V0/V)^2P0

D
P=(V/V0)^2P0

Solution

Resistance of bulb is R=V20P0
now P=V2R=V2P0V20=(VV0)2P0.
Hence, the correct option is (D)
Question 5
1 / -0

The semi-major axes of the orbits of Mercury and Mars in the astronomical units are 0.387 and 1.524 respectively. If the time period of Mercury is 0.241 year, then the time period of mars will be:

A
0.9 Year

B
0.19 Year

C
1.9 Year

D
2.9 Years

Solution

Tœa1.5
\(\operatorname{TIT} 2=(\operatorname{ata} 2) \operatorname{l.} 5 \rightarrow \operatorname{T1}=0.241 \times(1.5240 .387) 1.5=1.9 year\)
Hence, the correct option is (C)
Question 6
1 / -0

Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes. Consider one such fountain which is produced by a pipe of internal diameter 2cm2cm in which water flows at a rate 3ms−1. The enclosure has 100 holes each of diameter 0.05cm. The velocity of water coming out of the holes is (in ms−1).

A
0.48

B
96

C
24

D
48

Solution

Continuity Equation states that A1V1 = A2V2
where A is area of cross section and V is corresponding velocity.
Area is directly proportional to square of radius.
And in this case we have to multiply RHS by 100 because there are multiple outlets at that end.
So, our final equation reads like π(R1)2V1=100π(R2)2V2.
Substituting values, we get V2=48ms−1.
Hence, the correct option is (D)
Question 7
1 / -0

A rectangular tank is placed on a horizontal ground and is filled with water to a height H above the base. A small hole is made on one vertical side at a depth D below the level of water in the tank. The distance x from the bottom of the tank at which the water jet from the tank will hit the ground is.

A
2D(H−D)−−−−−−−−√

B
2DH−−−√

C
2D(H+D)−−−−−−−−√

D
12DH−−−√

Solution

The horizontal range of the liquid coming out a hole is equal to \(2 hh ^{\prime} \longrightarrow V ,\) where
\(h =\) depth of the hole below free surface of the liquid \(h ^{\prime}=\) height of the hole above the bottom of the tank
Here h=D and h'=H-D
So, range of the water stream.
x=2D(H−D)−−−−−−−−√
Question 8
1 / -0

A wooden block of weight 20 N is pushed along a rough horizontal surface at constant speed by a horizontal force of 10 N. Which of the following is the coefficient of static friction between the block and the surface? (μs>μk)

A
Zero

B
Less than 0.5

C
Equal to 0.5

D
Greater than 0.5

Solution

Let,
\(W\) is the weight of the block, \(F\) is force on block.
As per the problem,
\(W = mg =2 ON \ldots \ldots \ldots \ldots .( 1 )\)
\(N = mg =2 ON \ldots \ldots \ldots \ldots .(2)\)
where, variables have their usual meanings.
As it is moving with constant speed means zero acceleration.
\(W = F\)
\(F =\mu kN\)
\(IO =\mu k (20)\)
\(\mu k =0.5\)
Therefore, As the block is moving which implies it has overcome this frictional force giving us the coefficient of friction as \(\mu s >0.5.\)
Hence, the correct option is (D)
Question 9
1 / -0

A steady current is flowing in a circular coil of radius R, made up of a thin conducting wire. The magnetic field at the center of the loop is BL. Now, a circular loop of radius R/n is made from the same wire without changing its length, by unfolding and refolding the loop, and the same current is passed through it. If new magnetic field at the center of the coil is BC, then the ratio BL/BC is:

A
1 : n2

B
n1/2

C
n : 1

D
none of these

Solution

\(BL =\mu OI 2 \pi R\)
If the radius is \(R / n\), the number of turns will be n.
\(BC = n \mu O I 2 \pi( R / n )= n 2 \mu O I 2 \pi R\)
Hence, BLBC=1:n2
Hence, the correct option is (A)
Question 10
1 / -0

The graph, shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and y having same surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissive power (E) and absorptive power (a) of the two bodies:

A
Ex>Ey and ax

B
Exay

C
Ex>Ey and ax>ay

D
Ex

Solution

Power Emitter
so the more the power radiated will show more rapid decrease in temperature which is more for body x than y.
\(so , ex >ey\)
since the body which radiates more absorbs more so similarly,
\(a x>2 y\)
Hence, the correct option is (C)

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Physics Test - 11

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Physics Test - 11

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SHARING IS CARING

Question 1

1 : 1

1 : 5

5 : 1

25 : 1

Solution

Question 2

1 : 2

4 : 1

1 : 4

2 : 1

Solution

Question 3

200 mW

250 mW

100 mW

150 mW

Solution

Question 4

P=(V/V0)P0

P=(V0/V)P0

P=(V0/V)^2P0

P=(V/V0)^2P0

Solution

Question 5

0.9 Year

0.19 Year

1.9 Year

2.9 Years

Solution

Question 6

0.48

96

24

48

Solution

Question 7

2D(H−D)−−−−−−−−√

2DH−−−√

2D(H+D)−−−−−−−−√

12DH−−−√

Solution

Question 8

Zero

Less than 0.5

Equal to 0.5

Greater than 0.5

Solution

Question 9

1 : n2

n1/2

n : 1

none of these

Solution

Question 10

Ex>Ey and ax

Exay

Ex>Ey and ax>ay

Ex

Solution

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