Shear Stress
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(Neglect buoyancy due to air).
- 1.8×10−10 N
- 3.8×10−11 N
- 5.8×10−10 N
- 4×10−10 N
(i) the shearing stress develop on surface is 68.1 pa
(ii) the shearing strain develop on surface is 0.25
(iii) the shear modulus of material is 274.4NM2
only statement (i) and (iii) are correct
only statement (i) and (iii) are correct
all the statements are correct
only statement (i) and (ii) are correct
- pole strength
- magnetic moment
- intensity of magnetisation
- moment of inertia
- 4
The breaking stress of a material is 108N/m, density of material is 3000kg/m3
. Find the greatest
length of wire that could hang vertically without breaking. (Pause the video and solve without
looking at the solution).
A metal ring of initial radius and cross-sectional area is fitted onto a wooden disc of radius .if Youngs modulus of metal is then tension in the ring is:
[Take π=3.14 and answer upto two decimal places]
- 18∘
- 0.18∘
- 36∘
- 0.36∘
- (−π2−1):(π2+14):(3π4+12)
- (−π2):(π2):(3π4−12)
- −π2:π2:3π4
- (−π2+1):(π2+1):(3π4+12)
- (μ04π)Md3
- (μ04π)3Md3
- √3×(μ04π)Md3
- √5×(μ04π)Md3
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The diagram shows the change x in length of a thin uniform wire caused by the application of stress F, at two different temperatures T1 and T2. The variations shown suggest that:
T1<T2
T1>T2
T1=T2
None
The length of the metallic wire is when the tension in it is It is when the tension is . The original length of the wire will be:
- 1.75
- length of wire
- area of cross-section of wire
- Both (a) and (b)
- independent of length and area of cross-section
A spring with natural length has a tension when its length is and the tension is when its length is. The natural length of spring will be:
- 520 mm
- 620 mm
- 670 mm
- 580 mm
- The elongation of A is more than that of B.
- The elongation of B is more than that of A.
- The energy stored in B is more than that in A.
- The energy stored in A is more than that in B.
- increasing the thermal conductivity of outer layer
- decreasing the thermal conductivity of inner layer
- by increasing the thickness of the inner layer
- by decreasing the thickness of the outer layer
what is the dimensional formula of co-efficient of shear modulus?
(ii)If the magnet is turned through 90∘ , the resultant field at the location of the null point will be.
- 2√5×10−4 T
- 4√5×10−4 T
- 3√5×10−4 T
- 8√5×10−4 T
- Poisson's ratio
- Young's modulus
- Bulk modulus
- Rigidity modulus
Shearing stress at this plane, in terms of F, A and θ
- F sin θ
- F sin2 θA
- F sin θcos θA
- F cos2 θA
- 2×109 Nm−2
- 4×109 Nm−2
- 8×109 Nm−2
- 16×109 Nm−2