Physics Test 267

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Physics Test 267

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Question 1
4 / -1

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the volume of the sheet to correct significant figures.

A
0.0755 m³

B
0.1855 m³

C
0.0855 m³

D
0.0852 m³

Solution

Length , l = 4,234 m

Breadth, b = 1.005 m

Thickness , t = 2.01 x 10^-2 m

Volume = l x b x t

⇒ V = 4.234 x 1.005 x 0.0201 = 0.0855289 = 0.0855 m³ (significant figure = 3)
Question 2
4 / -1

Population of a town is reported as 157,900 (rounded to the nearest hundred). Which of the following statements is correct?

A
The population is 157900 ± 100

B
The population is 157900 ± 50

C
The population is 157900 ± 20

D
The population is 157900 ± 25
Solution
The number 157,900 ends with two zeros, which means it has been rounded to the nearest hundred.

When a number is rounded to the nearest hundred, the possible actual value lies within ±50 of the given number.

This means the actual population could be as low as:
- 157,850 (if rounded up to 157,900)
Or as high as:
- 157,949 (if rounded down to 157,900)
But in standard approximation rules:
- Lower limit = 157,900 – 50 = 157,850
- Upper limit = 157,900 + 50 = 157,950
So, the maximum possible error is ±50.
Question 3
4 / -1

The product of energy and time is called action. The dimensional formula for action is same as that for

A
Power

B
Angular energy

C
Force × velocity

D
Impulse × distance

Solution

Action: Energy × Time = (M L² T⁻²) × (T) = M L² T⁻¹

Power: M L² T⁻³

Angular Momentum: M L² T⁻¹

Force × Velocity: (M L T⁻²) × (L T⁻¹) = M L² T⁻³

Impulse × Distance: (M L T⁻¹) × (L) = M L² T⁻¹

The dimensional formula for Action matches that of Impulse × Distance, which is M L² T⁻¹.

Therefore, the answer is D.
Question 4
4 / -1

The Van der Waals equation for 1 mole of a real gas is

where P is the pressure, V is the volume, T is the absolute temperature, R is the molar gas constant and a and b are Van der Waal constants. The dimensions of a are the same as those of

A
PV

B
PV²

C
P²V

D
P/V

Solution

The image provides a solution for finding the dimensions of a in the Van der Waals equation for 1 mole of a real gas:
Question 5
4 / -1

A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

A
4v²/5g

B
4g/5v²

C
4v³/5g²

D
4v/5g²

Solution
Question 6
4 / -1

The range R of projectile is same when its maximum height are h₁ and h₂. What is the relation between R, h₁ and h₂?

A

B

C

D

Solution

Thus, the range R for angles θ and 90° - θ is the same, and the relationship between the maximum heights at these two angles leads to this conclusion.
Question 7
4 / -1

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle q1; in the next 2 sec it rotates through an additional angle θ₂, the ratio θ₂/θ₁ is

A
1

B
2

C
3

D
5

Solution
Question 8
4 / -1

Two cars are moving in same direction with speed of 30 kmph. They are separated by a distance of 5 km. What is the speed of a car moving in opposite direction if it meets the two cars at an interval of 4 min?

A
60 kmph

B
5 kmph

C
30 kmph

D
45 kmph

Solution

Lets say the speed of second car be v. Now with respect to car 1, speed of car 2 is v - 30. Now if the cross each other at just after 4 min then the distance traveled by car 2 relative to car 1 is (v -30) x 4/60 and we know the distance travelled relatively is 5km

Thus we get

(v -30) x 4/60 = 5

V = 300/4 - 30

= 45 km/h
Question 9
4 / -1

A body starts from rest, the ratio of distances travelled by the body during 3^rd and 4^thseconds is :

A
7/5

B
5/7

C
7/3

D
3/7

Solution

The velocity after 2 sec = 2a and velocity after 3 sec = 3a

For some constant acceleration a,

Now distance travelled in third second, s₃ = 2a + ½ a

= 5/2 a

Similarly distance travelled in fourth second, s₄= 3a + ½ a

= 7/2 a

Hence the reqiuired ratio is 5/7
Question 10
4 / -1

The dimensional formula for acceleration is

A
[LT²]

B
[LT^-2]

C
[L²T]

D
[L²T²]

Solution

Acceleration (a) = Velocity × Time^-1 . . . . . (1)

Since, Velocity (v) = Displacement × Time^-1

∴ The dimensional formula of velocity is [M⁰ L¹ T^-1] . . . . (2)

On substituting equation (2) in equation (1) we get,

Acceleration = Velocity × Time^-1

Or, a = [M⁰ L¹ T^-1] × [T]^-1 = [M⁰ L¹ T^-2]

Therefore, acceleration is dimensionally represented as [M⁰ L¹ T^-2].