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The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of

1:2

2:1

4:1

1:1

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Explanation

Given, Y_steel=2Y_brass and L_s=L_b and A_s=A_b

such that ΔL_s=ΔL_b

As we know,Young's modulus

Y=stress/strain=W/A/ΔL/L

i.e. W_s/W_b=Y_s/Y_b=2Y_b/Y_b=2/1

=> 2:1

Thus, weight added to the steel and brass wires must be in the ratio of 2:1

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Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Δl. Which of the following graphs is a straight line?

Δl versues 1/l

Δl versus l²

Δl versus 1/l²

Δl versus l

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Explanation

For Hooke's law, Young Modulus

y=F/A/Δl/l

=F x l/Δl x A ...(i)

V = A x l=constant ...(ii)

From Eqs. (i) and (ii),

y=F x lxl / Δl x V

Δl=F/V x y x l²

=>Δl∝l²

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The following four wires are made of the same material. Which of then will have the largest extension when the same tension is applied?

Length=50cm, diameter=0.5mm

Length=100cm, diameter=1mm

Length=200cm, diameter=2mm

Length=300cm, diameter=3mm

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Explanation

(a) ΔL=FL/AY or ΔL∝L/d² [∴ A=πd²/4]

Therefore,ΔL will be maximum for that wire for which L/A is maximum.

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The Young's modulus of a wire of length L and radius r is $Y N / m^{2} .$ If the length and radius are reduced to L/2 and r/2, then its Young's modulus will be -

1:1

1:4

1:8

8:1

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Explanation

(a) Young's modulus of wire does not varies with dimension of wire. It is the property of given material.

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When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be -

0.5 cm

2 cm

4 cm

8 cm

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Explanation

(c)

$l = \frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^{2}} (F, L a n d Y a r e c o n s \tan t) \frac{l_{2}}{l_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(2)}^{2} = 4 \Rightarrow l_{2} = 4 l_{1} = 4 c m$

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The area of cross-section of a wire of length 1.1 metre is 1 $m m^{2} .$ It is loaded with 1 kg. If Young's modulus of copper is $1.1 \times 10^{11} N / m^{2},$ then the increase in length will be (If $g = 10 m / s^{2}$ )

0.01 mm

0.075 mm

0.1 mm

0.15 mm

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Explanation

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In CGS system, the Young's modulus of a steel wire is $2 \times 10^{12} d y n e / c m^{2}$ . To double the length of a wire of unit cross-section area, the force required is

$4 \times 10^{6}$ dynes

$2 \times 10^{12} d y n e s$

$2 \times 10^{12} n e w t o n s$

$2 \times 10^{8}$ dynes

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Explanation

(b) To double the length of wire,

Stress = Young's modulus

$∴ \frac{F}{A} = 2 \times 10^{12} \frac{d y n e}{c m^{2}}$

If A= $1$ then F= $2 \times 10^{12} d y n e$

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The material which practically does not show elastic after effect is

Copper

Rubber

Steel

Quartz

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Explanation

(d) When a elastic body is stretched and applied deforming force is removed then the body is expected to return to its original configuration instantaneously. But sometimes some materials take some time to return to its original configuration. This temporary delay in achieving its original configuration is termed as elastic after effect. This elastic after effect is very short for quartz fiber. But elastic after effect is more for glass.

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A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of radius 2R will be

F/2

F/4

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Explanation

If radius of wire is double then breaking force will become four times.

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The relationship between Young's modulus Y, Bulk modulus K and modulus of rigidity n is

$Y = \frac{9 n K}{n + 3 K}$

$\frac{9 Y K}{Y + 3 K}$

$Y = \frac{9 n K}{3 + K}$

$Y = \frac{3 n K}{9 n + K}$

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Explanation

(a)

$Y = 3 K (1 - 2 σ) \Rightarrow \frac{Y}{3 K} = 1 - 2 σ a n d Y = 2 n (1 + σ) \Rightarrow \frac{Y}{2 n} = 1 + σ O n d o i n g (1) + (2) \times 2 - \frac{Y}{3 K} + \frac{Y}{n} = 3 \Rightarrow \frac{Y (n + 3 K)}{3 n K} = 3 \Rightarrow Y = \frac{9 n K}{n + 3 K}$

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