- Refer to answer 26, page no. 305 (MTG CBSE
Champion Physics class 11)
OR
Refer to answer 20, page no. 336 (MTG CBSE
Champion Physics class 11) - (i) The resolved part of F along the normal is the
tensile force on this plane and the resolved part parallel
to the plane is the shearing force on the plane.
F sin q
F cos q
F F? Area of plane section = A sec qTensile stress =
Force
Area=F
Acos
secθ
θ=F
Acos^2 θ(ii) Shearing stress applied on the top face
So, F = F sin qShearing stress =Force
Area==F
AF
Asin
secsincosθ
θθθ(^) = F
2 A
sin 2q,
[∵ sin 2q = 2sinq cosq]
(iii) Tensile stress will be maximum when cos^2 q is
maximum i.e., cos q = 1 or q = 0°
(iv) Shearing stress will be maximum when
sin 2q is maximum i.e. sin 2q = 1
or 2q = 90° or q = 45°.
OR
Refer to answer 62, page no. 342 (MTG CBSE
Champion Physics class 11)
- The situation is shown in figure. Let AB be the
rod of length 2 m. Suppose a weight W is hung at C at
distance x from A. Let T 1 and T 2 be the tensions in the
steel and brass wire respectively.
x
WC
AT 1Steel
wireBrass
wire
T 22–x B(i) Stress in steel wire =T
A1
1,Stress in brass wire =T
A2
2As both the stresses are equal, so?T
AT
AT
TA
A1
12
21
21
201
021
2==or ==.
.
Now moments about C are equal as the system is
equilibrium
? T 1 x = T 2 (2–x) orT
Tx
x1
22
=−or1
2=^2 −x
x∵
T
T1
21
2=
x = 4 – 2x
? 3 x = 4 or x =^4
3= 1.33 m(ii) Now Y = Stress/Strain
? Strain = Stress/YStrain in steel wire =TA
Y11
1/Strain in brass wire =TA
Y22
2/NowT
AYT
AY1
112
22=?
T
TAY
AY1
211
222102
210201 20 10
02 10 10== 1××
××=−
−.
.cm Nm
cm NmAgain, T 1 x = T 2 (2–x) orT
Tx
x1
2=^2 −or 1 =^2 −x
x∵
T
T1
2= 1
? x = 2 –x
or 2x = 2 or x =1 m.
OR
Refer to answer 95, page no. 346 (MTG CBSE
Champion Physics class 11)
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