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3 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In an experiment to determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm. The error in the measurement of Young's modulus $(Δ Y)$ is found to be x * 10¹⁰Nm^-2. The value of x is _________.

(take g = 10 ms^-2)

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Vishal Baghel

Contributor-Level 10

L = 1 m

$Δ L = 0.4 * 1 0^{- 3} m$

$m = 1 k g$

d = 0.4 * 10^-3 m

$\frac{F}{A} = y \frac{Δ L}{L}$

$y = \frac{F L}{A Δ L} = \frac{(m g) * 1}{(\frac{π d^{2}}{4}) * 0.4 * 1 0^{- 3}}$

$Δ y = 0.1 * 0.199 * 1 0^{12} = 1.99 * 1 0^{10}$

= 1.99

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3 months ago

Mechanical Properties of Solids Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine

The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$ . It is $l_{2}$ when the tension is T₂. The original length of the wire will be

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Payal Gupta

Contributor-Level 10

Stress = Y * strain

$\frac{T_{1}}{A} = Y * - \frac{(l_{1} - l)}{l} (i)$

$\frac{T_{2}}{A} = Y * \frac{(l_{2} - l)}{l} (i i)$

$\Rightarrow$ $\frac{T_{1}}{T_{2}} = \frac{l_{1} - l}{l_{2} - l}$

$\Rightarrow l = \frac{T_{1} l_{2} - T_{2} l_{2}}{T_{1} - T_{2}} = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

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New question posted

3 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$ . It is $l_{2}$ when the tension is T₂. The original length of the wire will be

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New answer posted

4 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A closely wounded circular coil of radius 5cm produces a magnetic field of 37.68 * 104 T at its centre. The current through the coil is_ A. [Given, number of turns in the coil is 100 and = 3.14]

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Vishal Baghel

Contributor-Level 10

$B_{c e n t r e} = N \frac{μ_{0} i}{2 r}$

$\Rightarrow \frac{100 * 4 π * 1 0^{- 7} * i}{2 * 5 * 1 0^{- 2}} = 37.68 * 1 0^{- 4}$

$∴ i = \frac{37.68}{4 * 3.14} = 3 A$

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New answer posted

4 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The pressure P1 and density d1 of diatomic gas $(γ = \frac{7}{5})$ changes suddenly to P2(>P1) and d2 respectively during an adiabatic process. The temperature of the gas increases and becomes ________times of its initial temperature. (given $\frac{d_{2}}{d_{1}} = 32)$

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Vishal Baghel

Contributor-Level 10

For adiabatic process – PVY = const

$T_{1} {V_{1}}^{Y - 1} = T_{2} V_{2}^{Y - 1}$

$\frac{T_{2}}{T_{1}} = {(\frac{v_{1}}{v_{2}})}^{Y - 1} = {(\frac{d_{2}}{d_{1}})}^{Y - 1} = {(32)}^{(\frac{7}{5} - 1)} = {(32)}^{2 / 5}$

$= {(2)}^{2} = 4$

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4 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Eight

If the length of a wire is made double and radius is halved of its respective values. Then, the Young’s modulus of the material of wire will :

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4 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 12th Chapter Nine

Given below are two statements : One is labeled as Assertion (A) and the other is labeled as Reason (R).

Assertion (A) : Clothes containing oil grease statins cannot be cleaned by water wash.

Reason (R) : Because the angle of contact between the oil/grease and water is obtuse.

In the light of the above statements, choose the correct answer from the option given below.

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Vishal Baghel

Contributor-Level 10

Water molecule doesnâ€™t stick with oil/ grease drops due to obtuse angle of contact between then.

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5 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1cm. The two wires will have

(a) The same stress.

(b) Different stress.

(c) The same strain.

(d) Different strain.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) for steel wire Y_steel= stress/strain= $\frac{f / A}{s t a r i n}$

When F and A are same for both the wires . hence stress will be same for both the wire

(Strain)_steel= stress/Y_steel and strain_copper=stress/Y_copper

Y_{steel
$\neq$}Y_copper

hence they both have different starin

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5 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 11th Chapter Nine

For an ideal liquid

(a) The bulk modulus is infinite.

(b) The bulk modulus is zero.

(c) The shear modulus is infinite.

(d) The shear modulus is zero.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) An ideal liquid is not compressible

Hence $?$ V=0

Bulk modulus B= strss /volumetric strain= $\frac{\frac{F}{A}}{\frac{? V}{V}} = \infty$

Compressibility K= 1/B=1/ $\infty = 0$

As there is no tangential force exists. So shear strain =0

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New answer posted

5 months ago

Mechanical Properties of Solids Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Fig. 9.4). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. Y_Al=70 $* 10^{9} N m^{- 2}$ and Y_steel= 200 $* 10^{9} N m^{- 2}$

(a) Mass m should be suspended close to wire A to have equal stresses in both the wires

(b) Mass m should be suspended close to B to have equal stresses in both the wires

(c) Mass m should be suspended at the middle of the wires to have equal stresses in both the wires

(d) Mass m should be suspended close to wire A to have equal strain in both wires

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) Let mass m is placed at x from the end B respectively.

T_A and T_B be the tensions in wire A and wire B respectively.

For the rotational equilibrium of the system,

\sum τ = 0

T_Bx-T_A(l-x)=0

$\frac{T_{B}}{T_{A}}$ = $\frac{l - x}{x}$

Stress in wire A = S_A= $\frac{T_{A}}{a_{A}}$

Stress in wire B = S_B= $\frac{T_{B}}{a_{B}}$ where a are the area of wire

We know that a_B=2a_A

Now for equal stress

S_A=S_B

$\frac{T_{A}}{a_{A}} = \frac{T_{B}}{a_{B}}$

$\frac{T_{B}}{T_{A}} = 2$

$\frac{l - x}{x} = 2$

So x =l/3 and l-x= 2l/3

Hence mass m should placed to B.

For equal strain

Strain_A= Strain_B

$\frac{Y_{A}}{S_{A}} = \frac{Y_{B}}{S_{B}}$

$\frac{Y_{s t e e l}}{Y_{A l}} = \frac{T_{A}}{T_{B}} * \frac{a_{B}}{a_{A}} = \frac{x}{l - x} (\frac{2 a_{A}}{a_{A}})$

$\frac{200 * 10^{9}}{70 * 10^{9}} = \frac{x}{l - x}$

After solving we get x= x= 10l/17

l-x=l=10l/17=7l/17

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