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Mechanical Properties of Solids

140980 The bulk modulus of a spherical object is "B". If it is subjected to uniform pressure ' $p$ ', the fractional decrease in radius is

1

\frac{B}{3 p}

2

\frac{3 p}{B}

3

\frac{p}{3 B}

4

\frac{p}{B}

Explanation:

C We know that,
Bulk modulus $(B) = \frac{pV}{dV}$
$∵ V = \frac{4}{3} π r^{3}$
differentiating w.r.t. r, we get
$\frac{dV}{dr} = 4 π r^{2}$
or $dV = 4 π r^{2} dr$
Putting the value of $V$ and $dV$ in equation (i), we get
$\begin{array}{ll} \frac{F_{1}}{F_{2}} = \frac{Δ L_{1}}{Δ L_{2}} \\ \frac{m_{1} g}{m_{2} g} = \frac{Δ L_{1}}{Δ L_{2}} {∵ Δ L_{1} = (L_{1} - L) \\ \frac{1}{2} = \frac{(L_{1} - L)}{(L_{2} - L)} and Δ L_{2} = (L_{2} - L)} \\ L_{2} - L = 2 L_{1} - 2 L \\ 2 L - L = 2 L_{1} - L_{2} \\ L = 2 L_{1} - L_{2} \end{array}$

CG PET-2021

Mechanical Properties of Solids

140981 Consider a rod of length $1.0 m$ with a crosssectional area of $0.50 {cm}^{2}$ . The rod supports a 500-kg platform that hangs attached to the rod's lower end. What is elongation of the rod under the stress ignoring the weight of the rod? Consider the Young's modulus to be $10^{11} Pa$ and $g = 10 m / s^{2}$

1

2 mm

2

0.5 mm

3

1.5 mm

4

1 mm

Explanation:

D Given that, length of $rod (L) = 1.0 m = 1000$ $mm$
Cross-section area of $rod (A) = 0.50 {cm}^{2} = 0.50 \times 10^{2}$ ${mm}^{2}$
Rod support mass $(m) = 500 kg$
Weight $(W) = mg = 500 \times 10 = 5000 N$
Young modulus $(Y) = 10^{11} Pa = 10^{11} \times 10^{- 6} N / {mm}^{2}$
$Y = 10^{5} N / {mm}^{2}$
$g = 10 m / \sec^{2}$
Elongation of the rod $(Δ L)$
$Δ L = \frac{WL}{AY}$
$Δ L = \frac{5000 \times 1000}{50 \times 10^{5}}$
$Δ L = 1 mm$

Shift-I]

Mechanical Properties of Solids

140983 The Young's modulus of a rubber string of length $12 cm$ and density $1.5 {kgm}^{- 3}$ is $5 \times 10^{8} N$ . $m^{- 2}$ . When this string is suspended vertically, the increase in its length due to its own weight is ___ (take $g = 10 m . s^{- 2}$ )

1

2.16 \times 10^{- 10} m

2

9.6 \times 10^{- 11} m

3

9.6 \times 10^{- 3} m

4

2.16 \times 10^{- 3} m

Explanation:

A Given, density $(ρ) = 1.5 kg / m^{3}$ , length $(L) =$ $12 cm = 12 \times 10^{- 2} m$ , Young's modulus $(Y) = 5 \times 10^{8}$ $N / m^{2}$
The increase in length due to self weight given by -
$Δ L = \frac{ρ \times g \times L^{2}}{2 Y}$
$Δ L = \frac{{(12 \times 10^{- 2})}^{2} \times 1.5 \times 10}{2 \times 5 \times 10^{8}}$
$Δ L = \frac{144 \times 10^{- 4} \times 1.5 \times 10}{10^{9}}$
$Δ L = 216 \times 10^{- 12}$
$Δ L = 2.16 \times 10^{- 10}$

Shift-I]

Mechanical Properties of Solids

140985 A wire of length $L$ , area of cross-section $A$ is hanging from a fixed support. The length of the wire changes to $L_{1}$ when mass $M$ is suspended from its free end. The expression for Young's modulus is

1

\frac{Mg (L_{1} - L)}{AL}

2

\frac{MgL}{{AL}_{1}}

3

\frac{MgL}{A (L_{1} - L)}

4

\frac{MgL}{AL}

Explanation:

C Given that,
Length of wire $= L$
Cross section area $= A$
Changes in length $= (L_{1} - L)$
Stress $= Mg / A$
Strain $= \frac{Δ L}{L} = \frac{(L_{1} - L)}{L}$
As we know that,
Young's modulus $Y = \frac{Stress}{Strain}$
$= \frac{Mg / A}{(L_{1} - L) / L}$
$Y = \frac{MgL}{A (L_{1} - L)}$

NEET SEP-2020

Mechanical Properties of Solids

140980 The bulk modulus of a spherical object is "B". If it is subjected to uniform pressure ' $p$ ', the fractional decrease in radius is

1

\frac{B}{3 p}

2

\frac{3 p}{B}

3

\frac{p}{3 B}

4

\frac{p}{B}

Explanation:

C We know that,
Bulk modulus $(B) = \frac{pV}{dV}$
$∵ V = \frac{4}{3} π r^{3}$
differentiating w.r.t. r, we get
$\frac{dV}{dr} = 4 π r^{2}$
or $dV = 4 π r^{2} dr$
Putting the value of $V$ and $dV$ in equation (i), we get
$\begin{array}{ll} \frac{F_{1}}{F_{2}} = \frac{Δ L_{1}}{Δ L_{2}} \\ \frac{m_{1} g}{m_{2} g} = \frac{Δ L_{1}}{Δ L_{2}} {∵ Δ L_{1} = (L_{1} - L) \\ \frac{1}{2} = \frac{(L_{1} - L)}{(L_{2} - L)} and Δ L_{2} = (L_{2} - L)} \\ L_{2} - L = 2 L_{1} - 2 L \\ 2 L - L = 2 L_{1} - L_{2} \\ L = 2 L_{1} - L_{2} \end{array}$

CG PET-2021

Mechanical Properties of Solids

140981 Consider a rod of length $1.0 m$ with a crosssectional area of $0.50 {cm}^{2}$ . The rod supports a 500-kg platform that hangs attached to the rod's lower end. What is elongation of the rod under the stress ignoring the weight of the rod? Consider the Young's modulus to be $10^{11} Pa$ and $g = 10 m / s^{2}$

1

2 mm

2

0.5 mm

3

1.5 mm

4

1 mm

Explanation:

D Given that, length of $rod (L) = 1.0 m = 1000$ $mm$
Cross-section area of $rod (A) = 0.50 {cm}^{2} = 0.50 \times 10^{2}$ ${mm}^{2}$
Rod support mass $(m) = 500 kg$
Weight $(W) = mg = 500 \times 10 = 5000 N$
Young modulus $(Y) = 10^{11} Pa = 10^{11} \times 10^{- 6} N / {mm}^{2}$
$Y = 10^{5} N / {mm}^{2}$
$g = 10 m / \sec^{2}$
Elongation of the rod $(Δ L)$
$Δ L = \frac{WL}{AY}$
$Δ L = \frac{5000 \times 1000}{50 \times 10^{5}}$
$Δ L = 1 mm$

Shift-I]

Mechanical Properties of Solids

140983 The Young's modulus of a rubber string of length $12 cm$ and density $1.5 {kgm}^{- 3}$ is $5 \times 10^{8} N$ . $m^{- 2}$ . When this string is suspended vertically, the increase in its length due to its own weight is ___ (take $g = 10 m . s^{- 2}$ )

1

2.16 \times 10^{- 10} m

2

9.6 \times 10^{- 11} m

3

9.6 \times 10^{- 3} m

4

2.16 \times 10^{- 3} m

Explanation:

A Given, density $(ρ) = 1.5 kg / m^{3}$ , length $(L) =$ $12 cm = 12 \times 10^{- 2} m$ , Young's modulus $(Y) = 5 \times 10^{8}$ $N / m^{2}$
The increase in length due to self weight given by -
$Δ L = \frac{ρ \times g \times L^{2}}{2 Y}$
$Δ L = \frac{{(12 \times 10^{- 2})}^{2} \times 1.5 \times 10}{2 \times 5 \times 10^{8}}$
$Δ L = \frac{144 \times 10^{- 4} \times 1.5 \times 10}{10^{9}}$
$Δ L = 216 \times 10^{- 12}$
$Δ L = 2.16 \times 10^{- 10}$

Shift-I]

Mechanical Properties of Solids

140985 A wire of length $L$ , area of cross-section $A$ is hanging from a fixed support. The length of the wire changes to $L_{1}$ when mass $M$ is suspended from its free end. The expression for Young's modulus is

1

\frac{Mg (L_{1} - L)}{AL}

2

\frac{MgL}{{AL}_{1}}

3

\frac{MgL}{A (L_{1} - L)}

4

\frac{MgL}{AL}

Explanation:

C Given that,
Length of wire $= L$
Cross section area $= A$
Changes in length $= (L_{1} - L)$
Stress $= Mg / A$
Strain $= \frac{Δ L}{L} = \frac{(L_{1} - L)}{L}$
As we know that,
Young's modulus $Y = \frac{Stress}{Strain}$
$= \frac{Mg / A}{(L_{1} - L) / L}$
$Y = \frac{MgL}{A (L_{1} - L)}$

NEET SEP-2020

Mechanical Properties of Solids

140980 The bulk modulus of a spherical object is "B". If it is subjected to uniform pressure ' $p$ ', the fractional decrease in radius is

1

\frac{B}{3 p}

2

\frac{3 p}{B}

3

\frac{p}{3 B}

4

\frac{p}{B}

Explanation:

C We know that,
Bulk modulus $(B) = \frac{pV}{dV}$
$∵ V = \frac{4}{3} π r^{3}$
differentiating w.r.t. r, we get
$\frac{dV}{dr} = 4 π r^{2}$
or $dV = 4 π r^{2} dr$
Putting the value of $V$ and $dV$ in equation (i), we get
$\begin{array}{ll} \frac{F_{1}}{F_{2}} = \frac{Δ L_{1}}{Δ L_{2}} \\ \frac{m_{1} g}{m_{2} g} = \frac{Δ L_{1}}{Δ L_{2}} {∵ Δ L_{1} = (L_{1} - L) \\ \frac{1}{2} = \frac{(L_{1} - L)}{(L_{2} - L)} and Δ L_{2} = (L_{2} - L)} \\ L_{2} - L = 2 L_{1} - 2 L \\ 2 L - L = 2 L_{1} - L_{2} \\ L = 2 L_{1} - L_{2} \end{array}$

CG PET-2021

Mechanical Properties of Solids

140981 Consider a rod of length $1.0 m$ with a crosssectional area of $0.50 {cm}^{2}$ . The rod supports a 500-kg platform that hangs attached to the rod's lower end. What is elongation of the rod under the stress ignoring the weight of the rod? Consider the Young's modulus to be $10^{11} Pa$ and $g = 10 m / s^{2}$

1

2 mm

2

0.5 mm

3

1.5 mm

4

1 mm

Explanation:

D Given that, length of $rod (L) = 1.0 m = 1000$ $mm$
Cross-section area of $rod (A) = 0.50 {cm}^{2} = 0.50 \times 10^{2}$ ${mm}^{2}$
Rod support mass $(m) = 500 kg$
Weight $(W) = mg = 500 \times 10 = 5000 N$
Young modulus $(Y) = 10^{11} Pa = 10^{11} \times 10^{- 6} N / {mm}^{2}$
$Y = 10^{5} N / {mm}^{2}$
$g = 10 m / \sec^{2}$
Elongation of the rod $(Δ L)$
$Δ L = \frac{WL}{AY}$
$Δ L = \frac{5000 \times 1000}{50 \times 10^{5}}$
$Δ L = 1 mm$

Shift-I]

Mechanical Properties of Solids

140983 The Young's modulus of a rubber string of length $12 cm$ and density $1.5 {kgm}^{- 3}$ is $5 \times 10^{8} N$ . $m^{- 2}$ . When this string is suspended vertically, the increase in its length due to its own weight is ___ (take $g = 10 m . s^{- 2}$ )

1

2.16 \times 10^{- 10} m

2

9.6 \times 10^{- 11} m

3

9.6 \times 10^{- 3} m

4

2.16 \times 10^{- 3} m

Explanation:

A Given, density $(ρ) = 1.5 kg / m^{3}$ , length $(L) =$ $12 cm = 12 \times 10^{- 2} m$ , Young's modulus $(Y) = 5 \times 10^{8}$ $N / m^{2}$
The increase in length due to self weight given by -
$Δ L = \frac{ρ \times g \times L^{2}}{2 Y}$
$Δ L = \frac{{(12 \times 10^{- 2})}^{2} \times 1.5 \times 10}{2 \times 5 \times 10^{8}}$
$Δ L = \frac{144 \times 10^{- 4} \times 1.5 \times 10}{10^{9}}$
$Δ L = 216 \times 10^{- 12}$
$Δ L = 2.16 \times 10^{- 10}$

Shift-I]

Mechanical Properties of Solids

140985 A wire of length $L$ , area of cross-section $A$ is hanging from a fixed support. The length of the wire changes to $L_{1}$ when mass $M$ is suspended from its free end. The expression for Young's modulus is

1

\frac{Mg (L_{1} - L)}{AL}

2

\frac{MgL}{{AL}_{1}}

3

\frac{MgL}{A (L_{1} - L)}

4

\frac{MgL}{AL}

Explanation:

C Given that,
Length of wire $= L$
Cross section area $= A$
Changes in length $= (L_{1} - L)$
Stress $= Mg / A$
Strain $= \frac{Δ L}{L} = \frac{(L_{1} - L)}{L}$
As we know that,
Young's modulus $Y = \frac{Stress}{Strain}$
$= \frac{Mg / A}{(L_{1} - L) / L}$
$Y = \frac{MgL}{A (L_{1} - L)}$

NEET SEP-2020

Mechanical Properties of Solids

140980 The bulk modulus of a spherical object is "B". If it is subjected to uniform pressure ' $p$ ', the fractional decrease in radius is

1

\frac{B}{3 p}

2

\frac{3 p}{B}

3

\frac{p}{3 B}

4

\frac{p}{B}

Explanation:

C We know that,
Bulk modulus $(B) = \frac{pV}{dV}$
$∵ V = \frac{4}{3} π r^{3}$
differentiating w.r.t. r, we get
$\frac{dV}{dr} = 4 π r^{2}$
or $dV = 4 π r^{2} dr$
Putting the value of $V$ and $dV$ in equation (i), we get
$\begin{array}{ll} \frac{F_{1}}{F_{2}} = \frac{Δ L_{1}}{Δ L_{2}} \\ \frac{m_{1} g}{m_{2} g} = \frac{Δ L_{1}}{Δ L_{2}} {∵ Δ L_{1} = (L_{1} - L) \\ \frac{1}{2} = \frac{(L_{1} - L)}{(L_{2} - L)} and Δ L_{2} = (L_{2} - L)} \\ L_{2} - L = 2 L_{1} - 2 L \\ 2 L - L = 2 L_{1} - L_{2} \\ L = 2 L_{1} - L_{2} \end{array}$

CG PET-2021

Mechanical Properties of Solids

140981 Consider a rod of length $1.0 m$ with a crosssectional area of $0.50 {cm}^{2}$ . The rod supports a 500-kg platform that hangs attached to the rod's lower end. What is elongation of the rod under the stress ignoring the weight of the rod? Consider the Young's modulus to be $10^{11} Pa$ and $g = 10 m / s^{2}$

1

2 mm

2

0.5 mm

3

1.5 mm

4

1 mm

Explanation:

D Given that, length of $rod (L) = 1.0 m = 1000$ $mm$
Cross-section area of $rod (A) = 0.50 {cm}^{2} = 0.50 \times 10^{2}$ ${mm}^{2}$
Rod support mass $(m) = 500 kg$
Weight $(W) = mg = 500 \times 10 = 5000 N$
Young modulus $(Y) = 10^{11} Pa = 10^{11} \times 10^{- 6} N / {mm}^{2}$
$Y = 10^{5} N / {mm}^{2}$
$g = 10 m / \sec^{2}$
Elongation of the rod $(Δ L)$
$Δ L = \frac{WL}{AY}$
$Δ L = \frac{5000 \times 1000}{50 \times 10^{5}}$
$Δ L = 1 mm$

Shift-I]

Mechanical Properties of Solids

140983 The Young's modulus of a rubber string of length $12 cm$ and density $1.5 {kgm}^{- 3}$ is $5 \times 10^{8} N$ . $m^{- 2}$ . When this string is suspended vertically, the increase in its length due to its own weight is ___ (take $g = 10 m . s^{- 2}$ )

1

2.16 \times 10^{- 10} m

2

9.6 \times 10^{- 11} m

3

9.6 \times 10^{- 3} m

4

2.16 \times 10^{- 3} m

Explanation:

A Given, density $(ρ) = 1.5 kg / m^{3}$ , length $(L) =$ $12 cm = 12 \times 10^{- 2} m$ , Young's modulus $(Y) = 5 \times 10^{8}$ $N / m^{2}$
The increase in length due to self weight given by -
$Δ L = \frac{ρ \times g \times L^{2}}{2 Y}$
$Δ L = \frac{{(12 \times 10^{- 2})}^{2} \times 1.5 \times 10}{2 \times 5 \times 10^{8}}$
$Δ L = \frac{144 \times 10^{- 4} \times 1.5 \times 10}{10^{9}}$
$Δ L = 216 \times 10^{- 12}$
$Δ L = 2.16 \times 10^{- 10}$

Shift-I]

Mechanical Properties of Solids

140985 A wire of length $L$ , area of cross-section $A$ is hanging from a fixed support. The length of the wire changes to $L_{1}$ when mass $M$ is suspended from its free end. The expression for Young's modulus is

1

\frac{Mg (L_{1} - L)}{AL}

2

\frac{MgL}{{AL}_{1}}

3

\frac{MgL}{A (L_{1} - L)}

4

\frac{MgL}{AL}

Explanation:

C Given that,
Length of wire $= L$
Cross section area $= A$
Changes in length $= (L_{1} - L)$
Stress $= Mg / A$
Strain $= \frac{Δ L}{L} = \frac{(L_{1} - L)}{L}$
As we know that,
Young's modulus $Y = \frac{Stress}{Strain}$
$= \frac{Mg / A}{(L_{1} - L) / L}$
$Y = \frac{MgL}{A (L_{1} - L)}$

NEET SEP-2020

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