Physics Test

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Physics Test - 1

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Question 1
1 / -0

A ball falls on surface from \(10 \mathrm{~m}\) height and rebounds to \(2.5 \mathrm{~m}\). If duration of contact with floor is \(0.01\) sec then average acceleration during contact is ___ \(\mathrm{ms}^{-2}\).

A
707

B
1400

C
1700

D
1800

Solution

We know velocity of the ball when dropped from a height \(\mathrm{h}=10 \mathrm{~m}\)
\(\Rightarrow \mathrm{v}_{1}=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 10}=10 \sqrt{2}\)
And velocity of the ball when rebounds to a height of \(\mathrm{h}=2.5 \mathrm{~m}\)
\(\Rightarrow\mathrm{v}_{2}=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 2.5}=5 \sqrt{2}\)
Now we know acceleration \(\mathrm{a}=\frac{\mathrm{v}_{1}-\mathrm{v}_{2}}{\mathrm{t}}\)
Here \(t=0.01 \mathrm{~s}\)
\(\Rightarrow a=\frac{10 \sqrt{2}-5 \sqrt{2}}{0.01}=707 \mathrm{ms}^{-2}\)
Question 2
1 / -0

For the given uniform square lamina ABCD whose centre is O,

A
I_AC = I_EF

B
I_AC =√2 I_EF

C
√2 I_AC = I_EF

D
I_AD= 3I_EF

Solution

Using perpendicular axis theorem:
2I_AC= I₀, where I₀ is the moment of inertia about an axis perpendicular to the plane of the lamina and passing through O.
Both diagonals are perpendicular to each other.
Further, 2I_EF = Io
Two medians (lines dividing the sides of the square equally) are also perpendicular to each other, also perpendicular to the axis passing through the centre and perpendicular to the plane of lamina.
Question 3
1 / -0

Three particles, each of mass 'm' g, are situated at the vertices of an equilateral triangle ABC of side l cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB in the plane of ABC in gram-cm² unit will be

A
(3/4) ml²

B
2 ml²

C
(5/4) ml²

D
(3/2) ml²

Solution
Question 4
1 / -0

Three weights w, 2w and 3w are connected to an identical spring suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The position of the weight from the rod are such that

A
3w will be farthest

B
w will be farthest

C
all will be at the same distance

D
2w will be farthest

Solution
Question 5
1 / -0

A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time, a viscous fluid of mass 'm' is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period

A
decreases continuously

B
decreases initially and increases again

C
remains unaltered

D
increases continuously

Solution

According to law of conservation of momentum,
l_ω = constant
When viscous fluid of mass m is dropped and start spreading out then its moment of inertia increases and angular velocity decreases. But when it start falling then its moment of inertia again starts decreasing and angular velocity increases.
Question 6
1 / -0

What is the dimension of L/RCV, where L, R, C and V have their usual meanings?

A
M¹L¹A^-2T⁰

B
M⁰L^-1A^-1F¹

C
M¹L⁰A^-1 T⁰

D
M⁰L ⁰A^-1T⁰

Solution
Question 7
1 / -0

In the following figure, the height of the object is H₁ = + 2.5 cm. What is the height of the image of H₁?

A
- 5 cm

B
+ 5 cm

C
+ 7.5 cm

D
- 7.5 cm

Solution

Since the given mirror is a concave mirror, the image formed will be inverted and double the size of the object.
Height of the object = 2.5 cm
Height of the image = – 5 cm; -ve sign indicates that the image is inverted.
Question 8
1 / -0

A convex lens made up of a material of refractive index μ₁ is immersed in a medium of refractive index μ₂ as shown in the figure. The relation between μ₁ and μ₂ is

A
μ₁ < μ₂

B
μ₁ > μ₂

C
μ₁ = μ₂

D

Solution
Question 9
1 / -0

A Carnot engine works as a refrigerator between 250 K and 300 K. If it receives 750 calories of heat from the reservoir at the lower temperature, then the amount of heat rejected at the higher temperature is

A
900 cal

B
625 cal

C
750 cal

D
1000 cal

Solution
Question 10
1 / -0

F₁ and F₂ are focal lengths of the objective and the eyepiece of a telescope, respectively. The angular magnification for this telescope is equal to

A
\(\frac{F_{1}}{F_{2}}\)

B
\(\frac{F_{2}}{F_{1}}\)

C
\(\frac{F_{1} F_{2}}{F_{1}+F_{2}}\)

D
\(\frac{F_{1}+F_{2}}{F_{1} F_{2}}\)

Solution

In normal adjustment of the telescope, \(|m|=\frac{f_{0}}{f_{e}}=\frac{F_{1}}{F_{2}}\)