Class 11 Medical Physics - Mechanical Properties Of Solids

A rubber tube is longer when suspended vertically than when placed horizontally on a table. Explain?

When the rubber tube is suspended vertically then it get a force which elongate it. As from Hook's law we know,

$strain \propto stress$

So, on applying stress it gets elongated.But when it placed horizontally,it does not get any stress to be elongated.

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Mass of the big structure, M $=$ 50,000 kg
Inner radius of the column, r $=$ 30 cm $=$ 0.3 m
Outer radius of the column, R $=$ 60 cm $=$ 0.6 m
Young’s modulus of steel, $Y=2\times 10^{11}Pa$

Total force exerted, $F=Mg=50000\times 9.8N$

Stress = Force exerted on a single column $= 50000 \times \frac{9.8}{4} = 122500 N$
Young’s modulus, $Y=\frac{Stress}{Strain}$

Strain $=\frac{(F/A)}{Y}$
Where,
Area, $A=\pi(R^2-r^2)=\pi((0.6)^2-(0.3)^2)$

Strain $=122500/[\pi((0.6)^2-(0.3)^2)\times 2\times 10^{11}]=7.22\times 10^{-7}$

Hence, the compressional strain of each column is $7.22 \times 10^{-7}$ .

A uniform steel rod of $5{ mm }^{ 2 }$ cross section is heated from $0^\circ C$ to $25^\circ C$ . Calculate the force which must be exerted to prevent it from expanding. Also calculate strain.
( $\alpha$ for steel $=12\times { { 10 }^{ -6 } }/{^\circ C }$ and $\gamma$ for steel $=20\times { 10 }^{ 10 }{ N }/{ { m }^{ 2 } }$ )

$F={ \gamma }_{ steel }\times { \alpha }_{ steel }\times \Delta T\times A$

$=20\times { 10 }^{ 10 }\times 12\times { 10 }^{ -6 }\times 25\times 5\times { 10 }^{ -6 }$

$=30000\times { 10 }^{ 10 }\times { 10 }^{ -12 }$

$=300N$

Strain

$={ \alpha }_{ steel }\times \Delta T$

$=12\times { 10 }^{ -6 }\times 25$

$=300\times { 10 }^{ -6 }$

$=3\times { 10 }^{ -4 }$ .

A steel wire of diameter $1\times { 10 }^{ -3 }m$ is stretched by a force of $20N$ . Calculate the strain energy per unit volume. ( $Y$ steel $=2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$ )

Given

$d=1\times { 10 }^{ -3 }m$ ,

$F=20N$

$Y =2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$ ,

$r=\dfrac { d }{ 2 } =0.5\times { 10 }^{ -3 }m=5\times { 10 }^{ -4 }m$

$\therefore$ Cross sectional area

$A=\pi { r }^{ 2 }=3.142\times 25\times { 10 }^{ -8 }{ m }^{ 2 }$

Strain energy per unit volume of the wire

$=\dfrac { 1 }{ 2Y } { \left( stress \right) }^{ 2 }$

$=\dfrac { { \left( { F }/{ A } \right) }^{ 2 } }{ 2Y }$

$=\dfrac { { F }^{ 2 } }{ 2{ A }^{ 2 }Y }$

$=\dfrac { { \left( 20 \right) }^{ 2 } }{ 2\times { \left( 3.142\times 25\times { 10 }^{ -8 } \right) }^{ 2 }\times 2\times { 10 }^{ 11 } }$

$=1622{ J }/{ { m }^{ 3 } }$ .

Find the increase in pressure required to decrease the volume of a water sample by $0.01\%$ . Bulk modulus of water $=2.1\times 10^9N m^{-2}$ .

bulk modulus of the water is

$B=2.1\times { 10 }^{ 9 }N{ m }^{ -2 }$

in order to increase the volume(v) and water sample by 0.01% increasing the pressure P

$\dfrac { V\times 0.01 }{ 100 } =\triangle V$

$\dfrac { \triangle V }{ V } ={ 10 }^{ -4 }$

$B=\dfrac { PV }{ \triangle V }$

$=2.1\times { 10 }^{ 9 }\times { 10 }^{ -4 }$

$=2.1\times { 10 }^{ 5 }\dfrac { N }{ m^2 }$

A stone of mass $1\ Kg$ is attached to one end of a string of length $1\ m$ and breaking strength $500\N$ , and is whirled in a horizontal circle on a frictionless table top. The other end of the string is kept fixed. Find the maximum speed the stone can attain, without breaking the string.

1312601_1135368_ans_059082621d024dceb0da4d244507bef0.jpg

What is Plasticity?

Plasticity is the property of a body by virtue of which it does not oppose the deforming force applied and the body is deformed permanently.

The pressure at a point varies from $99980\ Pa$ to $100020\ Pa$ , due to a simple harmonic sound wave. The amplitude and wavelength of the wave are $5\times{10}^{-6 }m$ and $40\ cm$ , respectively. Find the bulk modulus of air.

1310784_1138247_ans_67d2099dfc6a401aa99fa8e51eed4ef8.jpg

A solid copper cube has an edge length of $85.5$ cm. How much stress must be applied to the cube to reduce the edge length to $85.0$ cm? The bulk modulus of copper is $1.4 \times 10^{11} N/m^2$

The cube has side length l and volume

$V = l^3$ . We use

$p=B\Delta V/V$ for the pressure p.

We note that:

$\dfrac{\Delta V}{V}=\dfrac{\Delta l^3}{l^3}=\dfrac{(\Delta l+l)^3-l^3}{l^3}=\dfrac{3\Delta l}{l}$
Thus, the pressure required is

$p=\dfrac{3B\Delta l}{l}=\dfrac{3(1.4 \times10^{11} N/m^2 )(85.5cm -85.0cm)}{85.5cm}=2.4\times10^9 \,N/m^2$

Compute the fractional change in volume of a glass slab , when subjected to hydraulic pressure of 10 atm .

Hydraulic pressure exerted = 10 atm

$10 \times 1.013 \times 10^{5} Pa$

$B =\dfrac{P.V}{\Delta V}$

$\dfrac{\Delta V}{V}$ is the fractional change in volume

$\dfrac{\Delta V}{V} = \dfrac{P}{B} = \dfrac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$

$= 2,73 \times 10^{-5}$

A steel cable with a radius of 1.5 cm support a chair lift at a ski area . If the maximum stress is not exceed $10 ^{8}N m^{2}$ , What is the maximum load the cable can support ?

r =n radius of cable = 1.5 cm

$A = \pi (0.015)^{2} m^{2}$

$max stress = 10^{8} Nm^{2}$

Let F be the maximum force .

$Stress = \dfrac{F}{A}$

$\therefore F = max stress \times area.$

$= 10^{8} \times\pi \times(0.015)^{2}$

$= 7.065 \times 10^{4} N$

= 70, 650 N

$\therefore$ The maximum load is 70,650 N

Compute the bulk modulus of water from the following data : Intial Volume = 100 . 0 liter .Pressure increase = 100.0 atm $\left ( 1 atm = 1.013 \times 10^{5} Pa \right ),$ Final Volume = 1000.5 litre .Compare the bulk modulus of water with that of air (at constant temperature ) . Explain in simple terms why the ratio is so large .

$V_{F} = final volume = 100. 5 I$

$100 . 5 \times 10^{-3} m^{3}$

$V_{i} = initial volume = 100 \times 10^{-3} m^{3}$

$\Delta V = V_{F} - V_{i} = 0.5 \times 10^{-3}m^{3}$

$= 5 \times 10^{-4} m^{3}$

$\Delta P = 100 atm = 100 \times 1.013 \times 10^{5}Pa$

A cube of side 3 cm Is subjected to shearing force .The toop of cube is sheared through 0.012 cm with respect to bottom face . If cube is made of a material of shear modulus $2.1 \times 10^{10} Nm^{-2}$ then find
Force applied .

Shearing stress

$= \dfrac{S . Force}{Area}$

Shearing Force

$= Shearing stress \times Area$

$8.4 \times 10^{7} \times (3 \times 10^{-1})^{2}$

$= 7.56 \times 10^{4}N$

75 . 6 KN .

A cube of 5 cm has its upper face displaced by 0.2 by a force of 8 N .Find Shear Modulus , Stress and Strain .

Shear Stress

$= \dfrac{F}{A} = \dfrac{8N}{25\times 10^{-4}m^{2}}$

$= 3200 Nm^{-2}$

$Shear Strain = \dfrac{\Delta l}{l}$

But

$\Delta I = 0.2 cm$

I = 5 cm

$Shear Strain = \dfrac{0.2 cm}{5 cm}$

$= 0.04$

Shear Modulus

$= \dfrac{Shear stress}{Shear strain}$

$= \dfrac{3200}{0.04}$

$= 80 \times 10^{3}Nm^{-2} or 80 kPa .$

Column I
(order of magnitude in Pa)

$T=10\times 10\ N=10^{2}\ N$

$Y=\dfrac{\dfrac{T}{A}}{\dfrac{\Delta L}{L}}$

$\Rightarrow \dfrac{Y}{1.5\times 10^{2}}=\dfrac{10^{2}}{\pi \left ( 0.2\times 10^{-3} \right )^{2}}$

$\Rightarrow Y=\dfrac{10^{4}\times 1.5}{\pi \left ( 0.2\times 10^{-3} \right )^{2}}=1.19\times 10^{11}\ N/m^2$

2)

$B=\dfrac{dp}{\dfrac{dv}{v}} ; Given: \dfrac{dv}{v}\times 100=0.1$

$\Rightarrow \dfrac{0.1}{10^{2}}\times 2.1\times 10^{9}=dp$

$\Rightarrow dp=2.1\times10^{6}$

3) Potential energy

$=v\times \dfrac{1}{2}\sigma \epsilon$

$=\dfrac{1}{2}\times \dfrac{\left ( 2\times 10^{8} \right )^{2}}{Y}\times 2\times 4\times 10^{-6}$

$=0.8$

$\Rightarrow Y=2\times 10^{11}$

$w=100\ rad/s$

$T=mw^{2}r$

$\Rightarrow T=10^{4}$

$=\dfrac{T}{A}$

$=\dfrac{10^{4}}{10^{-6}}$

$=10^{10}$

5)

$B$ =

$\dfrac{P}{\dfrac{\Delta V}{V}}=\dfrac{\rho }{\dfrac{\Delta \rho }{\rho }}$

$\Rightarrow \Delta \rho =\dfrac{P\rho }{B}$

$\Rightarrow (11.685-11.4)=\dfrac{2\times 10^{4}\times 11.4}{B}$

$\Rightarrow B = 8\times 10^{5}Pa$

$\Rightarrow$ B is of order of

$10^{5}\ Pa$

A smooth uniform string of natural length $l$ ,cross-sectional area $A$ and Young's modulus $Y$ is pulled aling its length by a force $F$ on a horizontal surface. If the elastic potential energy stored in the string is $U=\cfrac { { F }^{ 2 }l }{ xAY }$ . Find $x$ .

$\Delta l$ be the elongation of string, Young modulus

$\displaystyle Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l} \Rightarrow \Delta l=\frac{Fl}{AY}$
Elastic potential energy stored in the string is:

$\displaystyle U=\frac{1}{2}F.\Delta l=\frac{1}{2}F.(\frac{Fl}{AY})=\frac{F^2l}{2AY}$

$\implies x=2$

A ball of mass 'm' drops from a height 'h' which sticks to mass-less hanger after striking. Neglect overturning, find out the maximum extension in the rod. Assuming rod is massless.

Applying energy conservation
mg (h + x) =

$\displaystyle \frac{1}{2}\: \: \frac{k_{1}k_{2}}{k_{1}+k_{2}}x^{2}$
where

$\displaystyle k_{1}=\frac{A_{1}Y_{1}}{\ l _{1}}$

$\displaystyle k_{2}=\frac{A_{2}Y_{2}}{\ l _{2}}$
&

$\displaystyle k_{eq}=\frac{A_{1}A_{2}Y_{1}Y_{2}}{A_{1}Y_{1}\ l _{2}+A_{2}Y_{2}\ l _{2}}$

$\displaystyle k_{eq}x^{2}-2mgx-2mgh=0$

$\displaystyle x=\frac{2mg\pm \sqrt{4m^{2}g^{2}+8mghk_{eq}}}{2k_{eq}}$

$\displaystyle x_{max}=\frac{mg}{k_{eq}}+\sqrt{\frac{m^{2}g^{2}}{k_{eq}^{2}}+\frac{2mgh}{k_{eq}}}$

Find the depth of a lake at which density of water is 1% greater than at the surface. Given compressibility K = 50 x $\displaystyle 10^{-6}$ /atm

$\displaystyle B=\frac{\Delta p}{-\frac{\Delta V}{V}}$

$\displaystyle \frac{\Delta V}{V}=-\frac{\Delta P}{B}$
m =

$\displaystyle \rho$ V = const.

$\displaystyle d\rho V+dV.\rho =0$

$\displaystyle \Rightarrow$

$\displaystyle \frac{d\rho }{\rho}=-\frac{dV}{V}$
i.e.

$\displaystyle \frac{\Delta \rho }{\rho }=\frac{\Delta P}{B}$

$\displaystyle \Rightarrow$

$\displaystyle \frac{\Delta \rho }{\rho }=\frac{1}{100}$

$\displaystyle \frac{1}{100}=\frac{h\rho g}{B}$ [assuming

$\displaystyle \rho$ = const.]
h

$\displaystyle \rho$ g=

$\displaystyle \frac{B}{100}=\frac{1}{100K}$

$\Rightarrow$ h

$\displaystyle \rho$ g =

$\displaystyle \frac{1\times 10^{5}}{100\times 50\times 10^{-6}}$

h =

$\displaystyle \frac{10^{5}}{5000\times 10^{-6}\times 1000\times 10}$ =

$\displaystyle \frac{100\times 10^{3}}{50}=2$ km

A load $1$ kg produces a certain extension in the wire of length $3$ m and radius $5 \times 10^{-4}$ m. How much will be the lateral strain produced in the wire? $\left( { Y } = 7.48 \times 10 ^ { 10 } { N } / { m } ^ { 2 } \text { and } \sigma = 0.291 \right)$

1451155_1392386_ans_32e49cceed9544ffbf6f9f08304c48b1.jpg

The figure shows a $300$ kg cylinder that is horizontal. Three steel wires support the cylinder from a ceiling. Wires 1 and 3 are attached at the ends of the cylinder, and wire 2 is attached at the center. The wires each have a cross-sectional area of $2.00 \times 10^6 m^2$ . Initially (before the cylinder was put in place) wires 1 and 3 were $2.0000$ m long and wire 2 was $6.00$ mm longer than that. Now (with the cylinder in place) all three wires have been stretched. What are the tension in (a) wire 1 and (b) wire 2?

(a) Let

$d$ =

$0.00600$ m. In order to achieve the same final lengths, wires 1 and 3 must stretch an amount

$d$ more than wire 2 stretches:

$ΔL1 = ΔL3 = ΔL2 + d .$
Using the above relation ,we find relation between the forces :

$F_1=F_3=F_2+\dfrac{dAE}{L}$
Now,

$F_1 + F_3 + F_2 – mg = 0$ . Combining this with the previous relation leads to

$F_1= 1380 N=1.38\times10^3$ N
(b) Similarly,

$F_2$ = 180 N.

A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) Of equal lengths as shown in Fig. The cross - sectional areas of wires A ad B are $1.0 mm ^{2}$ and $2.0 mm^2$ respectively . At what point along the rod should a mass m be suspended in order to produce. (a) equal stresses and (b) equal strain in both steel and aluminium wires.
$\gamma_{steel}=2×10^{11} Nm^{-2}$ and $\gamma_{aluminium}=7×10^{10}Nm^{-2}$

For equal strain :
We know that , Strain

$= \dfrac{Stress}{Y} = \dfrac{F}{Ay}$ We require

$(strain)_{a} = (Strain)_{b}$

Cross -sectional area of wir A,
$a_1=1.0mm^2=1.0×10^{-6}m^2$

Cross -sectional area of wir B,
$a_2=2.0mm^2=2.0×10^{-6}m^2$

Young’s modulus for steel,
$Y_1=2×10^11Nm^{−2}$

Young’s modulus for aluminium,
$Y_2=7.0×10^10Nm^{−2}$

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
$Stress in the wire =\dfrac{ Force}{Area}=\dfrac{F}{a}$
If the two wires have equal stresses, then:
$\dfrac{F_1}{a_1}=\dfrac{F_2}{a_2}$
Where,
$F_1$ = Force exerted on the steel wire
$F_2$ = Force exerted on the aluminum wire
$\dfrac{F_1}{F_2}=\dfrac{a_1}{a_2}=\dfrac{1}{2}$ .... (i)
The situation is shown in the figure.

Taking torque about the point of suspension, we have:
$F_1y=F_2(1.05−y)$

$\dfrac{F_1}{F_2}=\dfrac{1.05−y}{y}$ ..... (ii)
Using equations (i) and (ii), we can write:
$\dfrac{1.05−y}{y}=\dfrac{1}{2}$
$2(1.05−y)=y$
$y=0.7m$
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young's modulus = $\dfrac{ Stress}{Strain}$

$Strain = \dfrac{Stress}{Young's modulus} =\dfrac{\dfrac{F}{a}}{Y}$
If the strain in the two wires is equal, then:
$\dfrac{\dfrac{F_1}{a_1}}{Y_1}=\dfrac{\dfrac{F_2}{a_2}}{Y_2}$
$\dfrac{F_1}{F_2}=\dfrac{a_1Y_1}{a_2Y_2}$

$\dfrac{a_1}{a_2}=\dfrac{1}{2}$

$\dfrac{F_1}{F_2}=\dfrac{1×2×10^{11}}{2×7×10^{10}}=\dfrac{10}{7}$ ..... (iii)

Taking torque about the point where mass m, is suspended at a distance
$y_1$ from the side where wire A attached, we get:
$F_1y_1=F_2(1.05–y_1)$

$\dfrac{F_1}{F_2}=\dfrac{1.05−y_1}{y_1}$ ..... (iv)

Using equations (iii) and (iv), we get:
$\dfrac{1.05−y_1}{y_1}=\dfrac{10}{7}$ 7(1.05−y_1)=10y_1y_1=0.432m$$
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.

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Mechanical Properties Of Solids - Class 11 Medical Physics - Extra Questions

A rubber tube is longer when suspended vertically than when placed horizontally on a table. Explain?

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

A steel wire of diameter $1\times { 10 }^{ -3 }m$ is stretched by a force of $20N$ . Calculate the strain energy per unit volume. ( $Y$ steel $=2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$ )

Find the increase in pressure required to decrease the volume of a water sample by $0.01\%$ . Bulk modulus of water $=2.1\times 10^9N m^{-2}$ .

A stone of mass $1\ Kg$ is attached to one end of a string of length $1\ m$ and breaking strength $500\N$ , and is whirled in a horizontal circle on a frictionless table top. The other end of the string is kept fixed. Find the maximum speed the stone can attain, without breaking the string.

What is Plasticity?

The pressure at a point varies from $99980\ Pa$ to $100020\ Pa$ , due to a simple harmonic sound wave. The amplitude and wavelength of the wave are $5\times{10}^{-6 }m$ and $40\ cm$ , respectively. Find the bulk modulus of air.

A solid copper cube has an edge length of $85.5$ cm. How much stress must be applied to the cube to reduce the edge length to $85.0$ cm? The bulk modulus of copper is $1.4 \times 10^{11} N/m^2$

Compute the fractional change in volume of a glass slab , when subjected to hydraulic pressure of 10 atm .

A steel cable with a radius of 1.5 cm support a chair lift at a ski area . If the maximum stress is not exceed $10 ^{8}N m^{2}$ , What is the maximum load the cable can support ?

A cube of side 3 cm Is subjected to shearing force .The toop of cube is sheared through 0.012 cm with respect to bottom face . If cube is made of a material of shear modulus $2.1 \times 10^{10} Nm^{-2}$ then find
Force applied .

A cube of 5 cm has its upper face displaced by 0.2 by a force of 8 N .Find Shear Modulus , Stress and Strain .

Column I
(order of magnitude in Pa)

A smooth uniform string of natural length $l$ ,cross-sectional area $A$ and Young's modulus $Y$ is pulled aling its length by a force $F$ on a horizontal surface. If the elastic potential energy stored in the string is $U=\cfrac { { F }^{ 2 }l }{ xAY }$ . Find $x$ .

A ball of mass 'm' drops from a height 'h' which sticks to mass-less hanger after striking. Neglect overturning, find out the maximum extension in the rod. Assuming rod is massless.

Find the depth of a lake at which density of water is 1% greater than at the surface. Given compressibility K = 50 x $\displaystyle 10^{-6}$ /atm

A load $1$ kg produces a certain extension in the wire of length $3$ m and radius $5 \times 10^{-4}$ m. How much will be the lateral strain produced in the wire? $\left( { Y } = 7.48 \times 10 ^ { 10 } { N } / { m } ^ { 2 } \text { and } \sigma = 0.291 \right)$

Class 11 Medical Physics Extra Questions

Mechanical Properties Of Solids - Class 11 Medical Physics - Extra Questions

A rubber tube is longer when suspended vertically than when placed horizontally on a table. Explain?

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

A steel wire of diameter 1×10−3m1\times { 10 }^{ -3 }m is stretched by a force of 20N20N. Calculate the strain energy per unit volume. (YY steel =2×1011N/m2=2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } })

Find the increase in pressure required to decrease the volume of a water sample by 0.01%0.01\%. Bulk modulus of water =2.1×109Nm−2=2.1\times 10^9N m^{-2}.

A stone of mass 1 Kg1\ Kg is attached to one end of a string of length 1 m1\ m and breaking strength 500\N500\N, and is whirled in a horizontal circle on a frictionless table top. The other end of the string is kept fixed. Find the maximum speed the stone can attain, without breaking the string.

What is Plasticity?

The pressure at a point varies from 99980 Pa99980\ Pa to 100020 Pa100020\ Pa, due to a simple harmonic sound wave. The amplitude and wavelength of the wave are 5×10−6m5\times{10}^{-6 }m and 40 cm40\ cm, respectively. Find the bulk modulus of air.

A solid copper cube has an edge length of 85.585.5 cm. How much stress must be applied to the cube to reduce the edge length to 85.085.0 cm? The bulk modulus of copper is 1.4×1011N/m21.4 \times 10^{11} N/m^2

Compute the fractional change in volume of a glass slab , when subjected to hydraulic pressure of 10 atm .

A steel cable with a radius of 1.5 cm support a chair lift at a ski area . If the maximum stress is not exceed 108Nm2 10 ^{8}N m^{2} , What is the maximum load the cable can support ?

A cube of side 3 cm Is subjected to shearing force .The toop of cube is sheared through 0.012 cm with respect to bottom face . If cube is made of a material of shear modulus 2.1×1010Nm−2 2.1 \times 10^{10} Nm^{-2} then findForce applied .

A cube of 5 cm has its upper face displaced by 0.2 by a force of 8 N .Find Shear Modulus , Stress and Strain .

Column I(order of magnitude in Pa)

A smooth uniform string of natural length ll,cross-sectional area AA and Young's modulus YY is pulled aling its length by a force FF on a horizontal surface. If the elastic potential energy stored in the string is U=F2lxAYU=\cfrac { { F }^{ 2 }l }{ xAY } . Find xx.

A ball of mass 'm' drops from a height 'h' which sticks to mass-less hanger after striking. Neglect overturning, find out the maximum extension in the rod. Assuming rod is massless.

Find the depth of a lake at which density of water is 1% greater than at the surface. Given compressibility K = 50 x 10−6\displaystyle 10^{-6} /atm

A load 11 kg produces a certain extension in the wire of length 33 m and radius 5×10−45 \times 10^{-4} m. How much will be the lateral strain produced in the wire? (Y=7.48×1010N/m2 and σ=0.291)\left( { Y } = 7.48 \times 10 ^ { 10 } { N } / { m } ^ { 2 } \text { and } \sigma = 0.291 \right)

Class 11 Medical Physics Extra Questions

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A steel wire of diameter $1\times { 10 }^{ -3 }m$ is stretched by a force of $20N$ . Calculate the strain energy per unit volume. ( $Y$ steel $=2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$ )

Find the increase in pressure required to decrease the volume of a water sample by $0.01\%$ . Bulk modulus of water $=2.1\times 10^9N m^{-2}$ .

A stone of mass $1\ Kg$ is attached to one end of a string of length $1\ m$ and breaking strength $500\N$ , and is whirled in a horizontal circle on a frictionless table top. The other end of the string is kept fixed. Find the maximum speed the stone can attain, without breaking the string.

The pressure at a point varies from $99980\ Pa$ to $100020\ Pa$ , due to a simple harmonic sound wave. The amplitude and wavelength of the wave are $5\times{10}^{-6 }m$ and $40\ cm$ , respectively. Find the bulk modulus of air.

A solid copper cube has an edge length of $85.5$ cm. How much stress must be applied to the cube to reduce the edge length to $85.0$ cm? The bulk modulus of copper is $1.4 \times 10^{11} N/m^2$

A steel cable with a radius of 1.5 cm support a chair lift at a ski area . If the maximum stress is not exceed $10 ^{8}N m^{2}$ , What is the maximum load the cable can support ?

A cube of side 3 cm Is subjected to shearing force .The toop of cube is sheared through 0.012 cm with respect to bottom face . If cube is made of a material of shear modulus $2.1 \times 10^{10} Nm^{-2}$ then find
Force applied .

Column I
(order of magnitude in Pa)

A smooth uniform string of natural length $l$ ,cross-sectional area $A$ and Young's modulus $Y$ is pulled aling its length by a force $F$ on a horizontal surface. If the elastic potential energy stored in the string is $U=\cfrac { { F }^{ 2 }l }{ xAY }$ . Find $x$ .

Find the depth of a lake at which density of water is 1% greater than at the surface. Given compressibility K = 50 x $\displaystyle 10^{-6}$ /atm

A load $1$ kg produces a certain extension in the wire of length $3$ m and radius $5 \times 10^{-4}$ m. How much will be the lateral strain produced in the wire? $\left( { Y } = 7.48 \times 10 ^ { 10 } { N } / { m } ^ { 2 } \text { and } \sigma = 0.291 \right)$