[Only for Dropper and XII batch]

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(3)Differentiate VPn=C w.r.to V using multiplication rule:VddVPn+PnddVV=0VnPn-1dPdV+Pn.1=0VnPn-1dP+Pn.dV=0nVPn-1dP=-Pn.dVdPdV=-PnnVPn-1=-PnVBulk Modulus=-dPdVV=--PnVV=Pn

Difficulty Level:

  • 29%
  • 34%
  • 28%
  • 11%

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(4)

U=12Fl=12σAl=12σllAl=12σεV=12×stress×strain×volume

Difficulty Level:

  • 10%
  • 10%
  • 10%
  • 72%

The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is 

(a)PB

(b) B3p

(c) 3pB

(d)p3B

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(d)The object is spherical and the bulk modulus is represented by B. It is the ratio of normal stress to the volumetric strain. 

Hence             B=F/AV/V

      VV=PBVV=PB

Hence p is applied pressure on the object and VV is

volume strain 

Fractional decreasesin volume 

       VV=3RR               So, V=43πR3

Volume of the sphere decreases due to the decrease in its radius.

Hence VV=3RR=PBRB=P3B

Difficulty Level:

  • 19%
  • 21%
  • 18%
  • 43%

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

(a)1:2

(b)2:1

(c)4:1

(d)1:1

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

Given, Ysteel=2Ybrass and Ls=Lb and As=Ab

such that ΔLs=ΔLb

As we know,Young's modulus

Y=stress/strain=W/A/ΔL/L



i.e. Ws/Wb=Ys/Yb=2Yb/Yb=2/1

=> 2:1

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

Difficulty Level:

  • 20%
  • 65%
  • 10%
  • 8%

The Young's modulus of a wire of length L and radius r is Y N/m2. If the length and radius are reduced to L/2 and r/2, then its Young's modulus will be -

(a) 1:1                           (b) 1:4

(c) 1:8                             (d) 8:1

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

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

Difficulty Level:

  • 68%
  • 19%
  • 11%
  • 5%

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

(a) F/2                                      (b) 2F 

(c) 4F                                       (d) F/4

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(c) Breaking Force µ Area of cross section of wire (πr2)

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

Difficulty Level:

  • 9%
  • 14%
  • 72%
  • 8%

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

(a) Y=9nKn+3K                                    (b) 9YKY+3K 

(c)  Y=9nK3+K                                     (d) Y=3nK9n+K         

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(a)

 Y=3K1-2σand Y=2n1+σEliminating σ we get Y=9nKn+3K

Difficulty Level:

  • 58%
  • 20%
  • 14%
  • 10%

If x longitudinal strain is produced in a wire of Young's modulus y, then energy stored in the material of the wire per unit volume is-

(a) yx2                                 (b) 2yx2  

(c) 12y2x                             (d) 12yx2 

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(d) Energy stored per unit volume =12×Stress×Strain

     =12×Young's modulus×Strain2=12×y×x2

Difficulty Level:

  • 7%
  • 6%
  • 15%
  • 74%

The Young's modulus of a rubber string 8 cm long and density 1.5 kg/m3 is 5×108 N/m2, is suspended on the ceiling in a room. The increase in length due to its own weight will be

(a) 9.6×10-5 m                           (b) 9.6×10-11m

(c) 9.6×10-3 m                           (d) 9.6 m

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(b) l=L2dg2Y=8×10-222×5×108=9.6×10-11m

Difficulty Level:

  • 13%
  • 67%
  • 17%
  • 5%

A and B are two wires of same material. The radius of A is twice that of B. They are stretched by the same load. Then the stress on B is

(a) Equal to that on A                  (b) Four times that on A

(c) Two times that on A               (d) Half that on A

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(b) 

Stress=forceArea  Stress1πr2SBSA=rArB2=22SB=4SA

Difficulty Level:

  • 13%
  • 71%
  • 8%
  • 10%