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Properties of Matter

JEE Physics · 167 questions · Page 17 of 17 · Click an option or "Show Solution" to reveal answer

Q161
Match List I with List II .tg .tg List I List II A. Surface tension I. kg m−1 s−1\mathrm{kg~m^{-1}~s^{-1}} B. Pressure II. kg ms−1\mathrm{kg~ms^{-1}} C. Viscosity III. kg m−1 s−2\mathrm{kg~m^{-1}~s^{-2}} D. Impulse IV. kg s−2\mathrm{kg~s^{-2}} Choose the correct answer from the options given below :
A A-IV, B-III, C-I, D-II
B A-IV, B-III, C-II, D-I
C A-I, B-I, C-III, D-IV
D A-III, B-IV, C-I, D-II
Correct Answer
Option A
Solution
 (A)  Surface Tension =Fℓ=MLT−2 L=ML0 T−2=kg s−2(IV)\begin{aligned} \text{ (A) } \text{ Surface Tension }=\frac{\mathrm{F}}{\ell} & =\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}=\mathrm{ML}^{0} \mathrm{~T}^{-2} \\\\ & =\mathrm{kg\,s}^{-2}(\mathrm{IV}) \end{aligned}
 (B) Pressure =F A=MLT−2 L2=kg m−1 s−2(III)\begin{aligned} & \text{ (B) Pressure }=\frac{F}{\mathrm{~A}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2} \\\\ & =\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-2}(\mathrm{III}) \end{aligned}

 (C) Viscosity =FA(dVdz)=MLT−2 L2(LT−1 L)=ML−1 T−1=kg m−1 s−1(I)\begin{aligned} \text{ (C) Viscosity } & =\dfrac{\mathrm{F}}{\mathrm{A}\left(\dfrac{\mathrm{dV}}{\mathrm{dz}}\right)}=\dfrac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2\left(\dfrac{\mathrm{LT}^{-1}}{\mathrm{~L}}\right)} \\\\ & =\mathrm{ML}^{-1} \mathrm{~T}^{-1}=\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-1}(\mathrm{I})\end{aligned}

 (D)  Impulse =∫Fdt=MLT−2×T=MLT−1=kg ms−1 (II) \begin{aligned} \text{ (D) } \text{ Impulse } & =\int F d t=\mathrm{MLT}^{-2} \times \mathrm{T} \\\\ & =\mathrm{MLT}^{-1}=\mathrm{kg\,ms}^{-1} \text{ (II) } \end{aligned}
Q162
Match List I with List II : .tg .tg LIST I LIST II A. A force that restores an elastic body of unit area to its original state I. Bulk modulus B. Two equal and opposite forces parallel to opposite faces II. Young's modulus C. Forces perpendicular everywhere to the surface per unit area same everywhere III. Stress D. Two equal and opposite forces perpendicular to opposite faces Choose the correct answer from the options given below : IV. Shear modulus Choose the correct answer from the options given below :
A (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
B (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
C (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
D (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Correct Answer
Option B
Solution

To match List I with List II, we must understand the definitions or concepts described in List I and associate them with the correct terms in List II.

A.

A force that restores an elastic body of unit area to its original state: This definition describes Stress, as stress is the force applied over a unit area of a material that results in a deformation.

Stress is involved in restoring the body to its original state.

B.

Two equal and opposite forces parallel to opposite faces: This is a description of Shear Modulus (also known as Modulus of Rigidity), which measures the material's response to shear stress (like forces applied parallel to one of its faces).

C.

Forces perpendicular everywhere to the surface per unit area same everywhere: This definition resembles the concept of Bulk Modulus, which is concerned with uniform pressure applied in all directions at points on the surface of a body and relates to volume change.

D.

Two equal and opposite forces perpendicular to opposite faces: This describes Young's Modulus (also known as Modulus of Elasticity), which involves tensile stress (force applied perpendicularly) leading to changes in length of a material.

Therefore, matching the lists results in the following: (A)-(III), because stress is the general concept applicable to the force per unit area described in restoring the body to its original state.

(B)-(IV), because shear modulus is specifically related to parallel forces applied on opposite faces.

(C)-(I), because bulk modulus deals with situations where force is applied uniformly over the surface of an object, affecting its volume.

(D)-(II), because Young's modulus applies to forces perpendicular to the faces, affecting the length of a material.

Hence, the correct option is Option B (A)-(III), (B)-(IV), (C)-(I), (D)-(II).

Q163
A load of mass M kg is suspended from a steel wire of length 2m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is:
A 5.0 mm
B zero
C 3.0 mm
D 4.0 mm
Correct Answer
Option C
Solution
FA=y.Δℓℓ{F \over A} = y.{{\Delta \ell } \over \ell }
Δℓ∝F\Delta \ell \propto F

. . .. (i) T == mg T == mg −- fB == mg −-

mρb.ρℓ.{m \over {{\rho _b}}}.{\rho _\ell }.

g

=(1−ρℓρb)= \left( {1 - {{{\rho _\ell }} \over {{\rho _b}}}} \right)

mg

=(1−28)= \left( {1 - {2 \over 8}} \right)

mg T' ==

34{3 \over 4}

mg From (i)

Δℓ′Δℓ=T′T=34{{\Delta \ell '} \over {\Delta \ell }} = {{T'} \over T} = {3 \over 4}
Δℓ′=34.Δℓ=3\Delta \ell ' = {3 \over 4}.\Delta \ell = 3

mm

Q164
If the terminal speed of a sphere of gold (density =19.5  kg/m3 = 19.5\,\,kg/{m^3}) is 0.20.2 m/sm/s in a viscous liquid (density =1.5  kg/m3 = 1.5\,\,kg/{m^3}, find the terminal speed of a sphere of silver (density =10.5  kg/m3 = 10.5\,\,kg/{m^3}) of the same size in the same liquid
A 0.40.4 m/sm/s
B 0.1330.133 m/sm/s
C 0.10.1 m/sm/s
D 0.20.2 m/sm/s
Correct Answer
Option C
Solution

Let Terminal velocity = vt Upward viscous force = downward weight of sphere

⇒6πηrvt=(43πr3)(ρ−σ)g\Rightarrow 6\pi \eta r{v_t} = \left( {{4 \over 3}\pi {r^3}} \right)\left( {\rho - \sigma } \right)g
⇒vt=2r2(ρ−σ)g9η\Rightarrow {v_t} = {{2{r^2}\left( {\rho - \sigma } \right)g} \over {9\eta }}

........ (1) where, ρ\rho = density of substance of a body σ\sigma = density of liquid Now let the terminal velocity of gold = vg and silver = vs.

From equation (1), we can write

vgvs=ρg−σρs−σ{{{v_g}} \over {{v_s}}} = {{{\rho _g} - \sigma } \over {{\rho _s} - \sigma }}
=19.5−1.510.5−1.5= {{19.5 - 1.5} \over {10.5 - 1.5}}

=

189{{18} \over 9}
=21= {2 \over 1}

∴\therefore

vs=vg2{v_s} = {{{v_g}} \over 2}

=

0.22{{0.2} \over 2}

= 0.1

Q165
A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3 R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal :
A L (1+29MgπYR2)\left( {1 + {2 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)
B L (1+13MgπYR2)\left( {1 + {1 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)
C L (1+19MgπYR2)\left( {1 + {1 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)
D L (1+23MgπYR2)\left( {1 + {2 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)
Correct Answer
Option B
Solution

Here r = 3R −-

2RL{{2R} \over L}

x ∴\therefore Extension in the wire of length dx, dl =

FdxAY{{Fdx} \over {AY}}

=

Mg dxπr2 Y{{Mg\,dx} \over {\pi {r^2}\,Y}}

=

Mg dxπ(3R−2RLx)2Y{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}

∴\therefore Change in wire length,

Δ\Delta

L =

∫0Ldl\int\limits_0^L {dl}

=

∫0LMg dxπ(3R−2RLx)2Y\int\limits_0^L {{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}}

=

MgπY∫0Ldx(3R−2RLx)2{{Mg} \over {\pi Y}}\int\limits_0^L {{{dx} \over {{{\left( {3R - {{2R} \over L}x} \right)}^2}}}}

=

MgπY[−1(3R−2RLx)×(−L2R)]0L{{Mg} \over {\pi Y}}\left[ { - {1 \over {\left( {3R - {{2R} \over L}x} \right)}} \times \left( { - {L \over {2R}}} \right)} \right]_0^L

=

MgπY{{Mg} \over {\pi Y}}
[(L2R2−L6R2)]\left[ {\left( {{L \over {2{R^2}}} - {L \over {6{R^2}}}} \right)} \right]

=

MgπY(2L6R2){{Mg} \over {\pi Y}}\left( {{{2L} \over {6{R^2}}}} \right)

=

MgL3πR2Y{{MgL} \over {3\pi {R^2}Y}}

∴\therefore The equilibrium extended length of the wire, = L +

Δ\Delta

L = L +

MgL3πR2Y{{MgL} \over {3\pi {R^2}Y}}

= L (1 +

13{1 \over 3}
MgπR2Y{{Mg} \over {\pi {R^2}Y}}

)

Q166
A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :-
A 0.8
B 0.7
C 0.6
D 0.5
Correct Answer
Option C
Solution

In 1st situation Vbρ\rhobg = Vsρ\rhowg

VsVb=ρbρw=45{{{V_s}} \over {{V_b}}} = {{{\rho _b}} \over {{\rho _w}}} = {4 \over 5}

...(i) Here Vb is volume of block Vs is submerged volume of block

ρb\rho _b

is density of block

ρw\rho _w

w is density of water & Let

ρo\rho _o

is density of oil Finally in equilibrium condition Vb

ρb\rho _b

g =

Vb2ρog+Vb2ρwg{{{V_b}} \over 2}{\rho _o}g + {{{V_b}} \over 2}{\rho _w}g
2ρb=ρ0+ρw2{\rho _b} = {\rho _0} + {\rho _w}
⇒ρoρw=35=0.6\Rightarrow {{{\rho _o}} \over {{\rho _w}}} = {3 \over 5} = 0.6
Q167
A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water = 103 kg/m3]
A 30.1 kg
B 87.5 kg
C 65.4 kg
D 46.3 kg
Correct Answer
Option B
Solution

Given

(50)3×30100×(1)×g=Mcubeg{\left( {50} \right)^3} \times {{30} \over {100}} \times \left( 1 \right) \times g = {M_{cube}}g

...(i) Let m mass should be placed Hence (50)3×\times (1) ×\times g = (Mcube + m)g …(ii) equation (ii) – equation (i) ⇒\Rightarrow mg = (50)3 × g(1 – 0.3) = 125 × 0.7 × 103 g ⇒\Rightarrow m = 87.5 kg

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