SAT (Stage-1) Mock Test

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SAT (Stage-1) Mock Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of densities of two bodies is 5 : 6 and their specific heats are in the ratio 3 : 5. The ratio of their thermal capacities per unit volume is

A
25 : 18

B
1 : 2

C
2 : 1

D
1 : 1

Solution

The densities of two substances are in the ratio 5:6 and their specific heats are in the ratio 3:5 respectively.
Thermal capacities per unit volume (T) = density(ρ) × specific heat(c)
Question 2
1 / -0

One gram of ice at 0°C is converted into steam at 100°C. The amount of heat required to do so is

A
80 cal

B
536 cal

C
616 cal

D
716 cal

Solution

Total heat required = Heat required to convert the ice into water + Heat required to raise the temperature of water from 0°C to 100°C + Heat required to convert water into steam

Latent heat of fusion, L_f = 80 cal/g, Specific heat of water, c = 1 cal/g°C,

Latent heat of vaporisation, L_v = 536 cal g

Q = mL_f + rncΔT + mL_v

Q = 1 × 80 + 1 × 1 × 100 + 1 × 536 = 716 cal or Q = 0.716 kcal
Question 3
1 / -0

In the circuit shown below, each of the four conductors is of resistance R. The potential difference between A and B is V. The current flowing between A and B is

A
V/R

B
2V/R

C
3V/R

D
4V/R

Solution

The circuit can be redrawn as shown in the figure.

Here all the four resistors are connected in parallel.

Equivalent Resistance of the network is R_eq = R/4

Current through network is I =
Question 4
1 / -0

All the resistances in the figure are in ohms. The effective resistance between points A and B is

A
6Ω

B
3Ω

C
9Ω

D
2Ω

Solution

In the given circuit,

The two resistances of 2Ω and 2Ω between 'Aab' are in series whose equivalent resistance is 4Ω. This is parallel with 4Ω resistance connected across Ab; hence, the equivalent resistance across Ab is 2Ω.

This network repeats itself thrice, which means the total resistance across AB is 2Ω.
Question 5
1 / -0

A vehicle is moving on a track as shown in figure. The weight of the vehicle is

A
maximum at A

B
maximum at B

C
maximum at C

D
same at A, B and C

Solution

When the vehicle is at point B, forces acting on it are

1. Weight Mg, acting downwards
2. Normal reaction R in upward direction.

Hence the centripetal force acting on the vehicle is

When the vehicle is at point C, forces acting on it are v:

3. Weight Mg, acting downwards
4. Normal reaction R in upward direction.

Hence the centripetal force acting on the vehicle is

When the vehicle is at point A, we have R = Mg
Hence the normal reaction is maximum when the vehicle is at point B, which means weight of the vehicle is maximum when it is at point B.
Question 6
1 / -0

In the given arrangement, the pulleys are fixed and ideal, the strings are light, m₁ > m₂, and S is a spring balance which is itself massless. The reading of S (in terms of mass) is

A
m₁ - m₂

B

C

D

Solution

m₁ > m₂

For the block m₁, we have

From equations 1 and 2, acceleration of the system,

Tension in the string will be,

The reading of the spring balance will be equal to the value of T.

So in terms of mass, reading of the S is
Question 7
1 / -0

If the radius of Earth shrinks by 1% and its mass remains the same, the acceleration due to gravity on the Earth`s surface would

A
increase by 1%

B
increase by 2%

C
decrease by 2%

D
decrease by 1%

Solution
Question 8
1 / -0

A wire carrying a current (I) is shaped as shown below. Section AB is a quarter circle of radius r. The magnetic field at C is directed

A
along the bisector of the angle ACB, away from AB

B
along the bisector of the angle ACB, towards AB

C
perpendicular to the plane of the paper, directed into the paper

D
at an angle π/4 to the plane of the paper

Solution

The magnetic field at point C due to the straight portion of the conductor is zero. Current in the quarter circle is clockwise. Using Right Hand Thumb rule "If the curling of the fingers of the right hand gives the sense of the current, then the direction of the thumb gives the direction of the magnetic field at the centre"; hence, the direction of the magnetic field at point C is directed downwards into the plane of paper.
Question 9
1 / -0

If the intensity ratio of two waves is 4 : 1, the ratio of their amplitudes will be

A
2 : 1

B
1 : 2

C
4 : 1

D
1 : 4

Solution

Intensity of the wave is directly proportional to the square of the amplitude of the wave.

Here, a is the amplitude of the wave.
Question 10
1 / -0

A man standing midway between two cliffs, claps his hands and starts hearing a series of echoes at intervals of one second. If the speed of sound in air is 340 ms^-1, the distance between the cliffs is

A
680 m

B
1700 m

C
340 m

D
1620 m

Solution

Let the distance between the two cliffs be d. Since, the man is standing midway between the two cliffs, then the distance of man from either end is d/2.
The distance travelled by sound (in producing an echo)

2 × d/2 = v × 1

⇒ d = 340 × 1

⇒ 340 m
Question 11
1 / -0

AB and CD are long straight conductors, distance d apart, carrying a current I. The magnetic field on BC due to the currents in AB and CD

A
is zero at all points

B
is zero only at its midpoint

C
has different magnitudes at different points

D
is maximum at its midpoint

Solution

The magnetic field on BC due to the currents in AB and CD has different magnitudes at different points.
According to Biot-Savart law:

where x is the variable distance from the long straight conducting wire(s).
Question 12
1 / -0

An object is initially at a distance of 500 cm from a plane mirror. If the mirror approaches the object at a speed of 2 cm/s, then after 6 seconds, the distance between the object and its image will be

A
476 cm

B
976 cm

C
226 cm

D
150 cm

Solution

Distance travelled by the mirror towards the object in 6 seconds,
d = vt = 2 × 6 =12 cm
Hence, distance between the object and the mirror = 500 - 12 = 488 cm
In case of the plane mirror, object distance = image distance
Hence, distance between the object and its image = 2 × 488 = 976 cm
Question 13
1 / -0

To obtain a magnified virtual image of an object by a convex lens of focal length f, the distance between the object and the lens should be

A
> 4f

B
between 2f and 4f

C
< f

D
> 6f

Solution

If the object is placed between the focus and the optical centre, then an enlarged, virtual and erect image will be formed on the same side as the object.
Question 14
1 / -0

Parsec is the unit of

A
time

B
distance

C
frequency

D
angular acceleration

Solution

The parsec (symbol: pc) is a unit of length used in astronomy. It is about 3.26 light-years, or just under 31 trillion (3.1 × 10¹³) kilometres (about 19 trillion miles).
Question 15
1 / -0

What is the SI unit of the universal gravitation constant 'G'?

A
Nm

B
Nm²

C
Nm²kg^-2

D
None of the above

Solution

Newton's law of gravitation is given by:

So, we can say that the unit of the universal gravitation constant is Nm²/kg²or Nm²kg^-2.