1. Two wires A and B are of the same material. Their lengths are in the ratio 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

1. 1. 2 : 1

   
   
   

   2. 1 : 4

   
   
   

   3. 1 : 8

   
   
   

   4. 8 : 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that Young's modulus
   

   Y =	F	×	L
   	πr²		l

   
   Since Y, F are same for both the wires, we have,
   

   	1		L1	=	1		L2
   	r1²		l1		r2²		l2

   
   

   or,	l1	=	r2² × L1	=	(D2/2) × L1	mv²
   	l2		r1² × L2		(D1/2) × L2

   
   

   or,	l1	=	D2² × L1	=	D2 L1	×	L2	=	1
   	l2		D1² × L2		(2D2)²		2L2		8

   
   So, l₁: l₂ = 1 : 8

   Correct Option: C

   We know that Young's modulus
   

   Y =	F	×	L
   	πr²		l

   
   Since Y, F are same for both the wires, we have,
   

   	1		L1	=	1		L2
   	r1²		l1		r2²		l2

   
   

   or,	l1	=	r2² × L1	=	(D2/2) × L1	mv²
   	l2		r1² × L2		(D1/2) × L2

   
   

   or,	l1	=	D2² × L1	=	D2 L1	×	L2	=	1
   	l2		D1² × L2		(2D2)²		2L2		8

   
   So, l₁: l₂ = 1 : 8

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2. ​If the ratio of diameters, lengths and Young’s modulus of steel and copper wires shown in the figure are p, q and s respectively, then the corresponding ratio of increase in their lengths would be
   
   
   

1. 1. 7q

      (5sp)

   
   
   

   2. 5q

      (7sp²)

   
   
   

   3. 7q

      (5sp²)

   
   
   

   4. 2q

      (5sp)

   
   
   

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1. View Hint View Answer Discuss in Forum

   From formula,
   

   Increase in length = ΔL =	FL	=	4FL
   	AY		πD²Y

   
   

   ΔLs

   =

   Fs

   Dc

   ²

   Yc

   Ls

   =

   7

   ×

   1

   ²

   1

   q

   ΔLc

   Fc

   Ds

   Yc

   Lc

   5

   p

   s

   
   

   =	7q
   	(5sp²)

   Correct Option: C

   From formula,
   

   Increase in length = ΔL =	FL	=	4FL
   	AY		πD²Y

   
   

   ΔLs

   =

   Fs

   Dc

   ²

   Yc

   Ls

   =

   7

   ×

   1

   ²

   1

   q

   ΔLc

   Fc

   Ds

   Yc

   Lc

   5

   p

   s

   
   

   =	7q
   	(5sp²)

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

1. 1. ​Length = 100 cm, diameter =  1 mm

   
   
   

   2. Length = 200 cm, diameter = 2 mm

   
   
   

   3. ​Length = 300 cm, diameter = 3 mm

   
   
   

   4. Length = 50 cm, diameter = 0.5 mm

   
   
   

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1. View Hint View Answer Discuss in Forum

   F = YA/L × l So, extension, l ∝ L/A ∝ L/D²  [∵ F and Y are constant]
   l₁ ∝ 100/l² ∝ 100 and l₂α (200/2²)∝ 50
   l₃∝ 300/3² ∝ 100/3 and l₄ ∝ (50/(1/4)) ∝ 200
   The ratio of L/D² is maximum for case (d).
   Hence, option (d) is correct.

   Correct Option: D

   F = YA/L × l So, extension, l ∝ L/A ∝ L/D²  [∵ F and Y are constant]
   l₁ ∝ 100/l² ∝ 100 and l₂α (200/2²)∝ 50
   l₃∝ 300/3² ∝ 100/3 and l₄ ∝ (50/(1/4)) ∝ 200
   The ratio of L/D² is maximum for case (d).
   Hence, option (d) is correct.

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4. 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 wi re is ‘∆l’. Which of the following graphs is a straight line?

1. 1. ∆l versus l/I

   
   
   

   2. ∆l versus l²

   
   
   

   3. ​∆l versus I/l²

   
   
   

   4. ∆l versus l

   
   
   

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1. View Hint View Answer Discuss in Forum

   As Y =	F		⇒ Δ l =
   	A			Fl
   	Al			AY
   	l

   
   

   But V = Al so A =	V
   	l

   
   

   Therefore ∆l =	Fl²	∝ l²
   	VY

   
   Hence graph of ∆l versus l2 will give a straight line.

   Correct Option: B

   As Y =	F		⇒ Δ l =
   	A			Fl
   	Al			AY
   	l

   
   

   But V = Al so A =	V
   	l

   
   

   Therefore ∆l =	Fl²	∝ l²
   	VY

   
   Hence graph of ∆l versus l2 will give a straight line.

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5. ​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 weights added to the steel and brass wires must be in the ratio of :

1. 1. 2 : 1

   
   
   

   2. 4 : 1

   
   
   

   3. 1 : 1

   
   
   

   4. 1 : 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Young's modulus Y =	W	.	l
   	A		Δl

   
   

   	W1	=	W2
   	Y1		Y2

   
   [ ∵ A, l, ∆l same for both brass and steel]
   
   

   	W1	=	W2	= 2
   	W1		Y2

   
   [Y_steel/Y_brass = 2  given]

   Correct Option: A

   Young's modulus Y =	W	.	l
   	A		Δl

   
   

   	W1	=	W2
   	Y1		Y2

   
   [ ∵ A, l, ∆l same for both brass and steel]
   
   

   	W1	=	W2	= 2
   	W1		Y2

   
   [Y_steel/Y_brass = 2  given]

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