In Searle’s experiment to find Young’s modulus the diameter of wire is measured as d = 0.05cm, length of wire is l= 125cm and when a weight, m = 20.0 kg is put, extension in wire was found to be 0.100cm. Find maximum permissible error in Young’s modulus (Y).Use : \(\frac{mg\iota }{(\pi /4)d^{2}x}\)

6.3%
5.3%
2.3%
1%

Solution

A steel wire of uniform cross-section of 1mm² is heated upto 50°C and clamped rigidly at its ends. If temperature of wire falls to 40°C, change in tension in the wire is (coefficient of linear expansion of steel is 1.1 × 10^–5/°C and Young’s modulus of elasticity of steel is 2 × 10¹¹N/m² )

22 N
44 N
88 N
88 × 10⁶ N

Solution

F=YA^Δl⁄_lYA α Δ θ

= 2 × 10¹¹× (1 × 10^–6) ×1.1 ×10^–5× (10) = 22 N

The upper end of a wire of diameter 12mm and length 1m is clamped and its other end is twisted through an angle of 30°.The angle of shear is

18°
0.18°
36°
0.36°

Solution

\(r\theta =\iota \phi \Rightarrow \phi =\frac{r\theta }{\iota }=\frac{6mm\times 30^{\circ}}{1m}=0.18^{\circ}\)

To break a wire, a force of 10⁶N/m² is required. If the density of the material is 3 × 10³kg/m³, then the length of the wire which will break by its own weight will be

34 m
30 m
300 m
3 m

Solution

\(L=\frac{S}{dg}=\frac{10^{6}}{3\times 10^{3}\times 10}=\frac{100}{3}=34m\)

A rod of length land radius r is joined to a rod of length l/2 and radius r/2 of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of θ°, the twist angle at the joint will be

θ/4
θ/2
5θ/6
8θ/48

Solution

The diagram shows a force – extension graph for a rubber band. Consider the following statements :

I.It will be easier to compress this rubber than expand it

II.Rubber does not return to its original length after it is stretched

III.The rubber band will get heated if it is stretched and released

Which of these can be deduced from the graph?

III only
II and III
I and III
I only

Solution

The potential energy U between two atoms in a diatomic molecules as a function of the distance x between atoms has been shown in the figure. The atoms are

attracted when x lies between A and B and are repelled when x lies between B and C
attracted when x lies between B and C and are repelled when x lies between A and B
are attracted when they reach B from C
are repelled when they reach B from A

Solution

The atoms when brought from infinity are attracted due to interatomic electrostatic force of attraction. At point B, the potential energy is minimum and force of attraction is maximum. But if we bring atoms closer than x = B,force of repulsion between two nuclei starts and P.E.increases.

A rubber cord catapult has cross-sectional area 25 mm² and initial length of rubber cord is 10 cm. It is stretched to 5cm and then released to project a missile of mass 5 gm.Taking r_ubber Y = 5 × 10⁸ N/m². Velocity of projected missile is

20 ms^–1
100 ms^–1
250 ms^–1
200 ms^–1

Solution

Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?

Solution

Two wires A and B are of the same material. Their lengths are in the ratio of 1: 2 and the diameter are in the ratio 2 : 1.If they are pulled by the same force, then increase in length will be in the ratio of

2 : 1
1 : 4
1 : 8
8 : 1

Solution

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