Properties Of Solids

A spherometer is an instrument specifically designed to measure the radius of curvature of a spherical surface and the thickness of small objects. Here's how it relates to the options provided in the question:

1. Radius of curvature of concave surface: A spherometer can be used to measure the radius of curvature of concave surfaces. It is one of the primary purposes of this instrument. The spherometer is placed on the surface, and the curvature is calculated based on the depression of the central screw relative to the outer legs.
2. Specific rotation of liquids: Specific rotation is a property of chiral compounds in solution, typically measured using a polarimeter, not a spherometer. Therefore, a spherometer cannot be used to measure this physical quantity.
3. Thickness of thin plates: By determining the difference in height between two surfaces (one where the object is present and one where it is absent), a spherometer can measure the thickness of thin plates effectively.
4. Radius of curvature of convex surface: Similar to concave surfaces, a spherometer can also measure the radius of curvature of convex surfaces. The technique remains largely the same, relying on the displacement of the central screw to signify the curvature.

Thus, specific rotation of liquids is the physical quantity that cannot be measured using a spherometer. The other options describe measurements that fall within the capabilities of a spherometer, which is primarily used for determining curvatures and thin thicknesses.

Correct Answer: Specific rotation of liquids

To solve this problem, we need to understand the motion of a small steel ball dropping into a viscous fluid like glycerine.

When the steel ball is dropped, it initially accelerates due to gravity. However, as it moves through the glycerine, it experiences an upward viscous drag force, which increases with velocity. Eventually, the ball reaches a constant velocity known as terminal velocity, where the net force on the ball becomes zero.

The forces acting on the ball are:

1. Gravitational force (downward)
2. Buoyant force (upward, due to glycerine)
3. Viscous drag force (upward, increases with velocity)

As these forces balance, the ball reaches its terminal velocity .

The velocity-time graph for this scenario will show:

• Initially, an accelerating slope as the ball speeds up due to gravity.
• Then, it will transition to a horizontal line once terminal velocity is reached, indicating constant velocity.

The correct graph representation is an initial steep slope that gradually levels off to a constant value, which matches with the provided image:

Hence, the correct answer is the option represented by the image above, where the velocity initially increases and then becomes constant after reaching terminal velocity.

Given:
Density of wire,
Breaking stress,
Acceleration due to gravity on the planet,

The breaking stress ( ) is given by:

Where is the tension, is the mass, and is the cross-sectional area.
Since (where is the length of the wire), we have:

Rearranging for :

Substituting the given values:

To solve this question, let's first understand the basic concept of Young's modulus and how it is affected by changes in the dimensions of a wire.

Concept: Young's modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. It is defined by the formula:

where:

• is the force applied,
• is the original length,
• is the original cross-sectional area,
• is the change in length.

Analysis:

The question provides that Young’s modulus of a material is for a wire of original length and cross-sectional area . If the length of the wire is doubled ( ) and the cross-sectional area is halved ( ), we need to determine how Young's modulus will change.

Since Young's modulus is a property of the material itself and is only dependent on the material properties (not on its dimensions), changing the dimensions of the wire will not affect Young's modulus. Young’s modulus remains constant for a given material under small deformations.

Conclusion:

Thus, even after doubling the length and halving the cross-sectional area of the wire, Young’s modulus remains . Therefore, the correct answer is:

To solve this problem, we use Bernoulli's equation for fluid flow. The principle states that in a steady flow, the sum of all forms of mechanical energy in a fluid along a streamline is constant. The equation in terms of pressure , fluid density , velocity , and gravitational potential energy is given by:



In our scenario, when the valve is closed, the water is at rest so . Also, assuming the heights and are the same, they cancel out. Therefore, the equation simplifies to:



We are given:

Substituting these into the simplified equation:



Solving for :



Assuming (density of water), the equation becomes:



Simplifying gives:





Therefore, . The value of is 50, which fits the expected range (50, 50).