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Properties of Matter

JEE Physics · 167 questions · Page 11 of 17 · Click an option or "Show Solution" to reveal answer

Q101
A water drop of radius 1 cm1 \mathrm{~cm} is broken into 729 equal droplets. If surface tension of water is 75 dyne/ cm\mathrm{cm}, then the gain in surface energy upto first decimal place will be : (Given π=3.14\pi=3.14 )
A 8.5×10−4 J8.5 \times 10^{-4} \mathrm{~J}
B 8.2×10−4 J8.2 \times 10^{-4} \mathrm{~J}
C 7.5×10−4 J7.5 \times 10^{-4} \mathrm{~J}
D 5.3×10−4 J5.3 \times 10^{-4} \mathrm{~J}
Correct Answer
Option C
Solution
729×43πr3=43πR3729 \times {4 \over 3}\pi {r^3} = {4 \over 3}\pi {R^3}
⇒R=9r\Rightarrow R = 9r

........ (1)

ΔU=S×ΔA\Delta U = S \times \Delta A

..... (2)

⇒ΔU=S×{−4πR2+729×4πr2}\Rightarrow \Delta U = S \times \{ - 4\pi {R^2} + 729 \times 4\pi {r^2}\}
=S×4π{729r2−81r2}= S \times 4\pi \{ 729{r^2} - 81{r^2}\}
=7.5×10−4 J= 7.5 \times {10^{ - 4}}\,J
Q102
The area of cross section of the rope used to lift a load by a crane is 2.5×10−4 m22.5 \times 10^{-4} \mathrm{~m}^{2}. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be : (take g=10 ms−2g=10 \,m s^{-2} )
A 6.25×10−4 m26.25\times 10^{-4} \mathrm{~m}^{2}
B 10×10−4 m210\times 10^{-4} \mathrm{~m}^{2}
C 1×10−4 m21\times 10^{-4} \mathrm{~m}^{2}
D 1.67×10−4 m21.67\times 10^{-4} \mathrm{~m}^{2}
Correct Answer
Option A
Solution

The relationship between stress (σ), force (F), and area (A) is given by :

σ=FA\sigma = \frac{F}{A}

In this context, the force is equal to the weight of the load, so we can substitute force with mass (m) times gravity (g) :

F=m⋅gF = m \cdot g

From this, we get the formula for the cross-sectional area required to support a given mass :

A=Fσ=m⋅gσA = \frac{F}{\sigma} = \frac{m \cdot g}{\sigma}

We can set up a proportionality relationship between the area for 10 metric tons (A₁₀) and the area for 25 metric tons (A₂₅) as follows :

A10A25=m10m25\frac{A_{10}}{A_{25}} = \frac{m_{10}}{m_{25}}

Using the given values : A10=2.5×10−4A_{10} = 2.5 \times 10^{-4} m², m10=10,000m_{10} = 10,000 kg, m25=25,000m_{25} = 25,000 kg, Solving for A25A_{25} :

A25=A10×(m25m10)=2.5×10−4 m2×(25,000 kg10,000 kg)=6.25×10−4 m2A_{25} = A_{10} \times \left(\frac{m_{25}}{m_{10}}\right) = 2.5 \times 10^{-4} \, \mathrm{m}^{2} \times \left(\frac{25,000 \, \mathrm{kg}}{10,000 \, \mathrm{kg}}\right) = 6.25 \times 10^{-4} \, \mathrm{m}^{2}

So, Option A (6.25×10−4 m2) (6.25 \times 10^{-4} \, \mathrm{m}^{2}) is the correct answer.

Q103
Two cylindrical vessels of equal cross-sectional area 16 cm216 \mathrm{~cm}^{2} contain water upto heights 100 cm100 \mathrm{~cm} and 150 cm150 \mathrm{~cm} respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take, density of water =103 kg/m3=10^{3} \mathrm{~kg} / \mathrm{m}^{3} and g=10 ms−2\mathrm{g}=10 \mathrm{~ms}^{-2} ] :
A 0.25 J
B 1 J
C 8 J
D 12 J
Correct Answer
Option B
Solution
A=16×10−4A = 16 \times {10^{ - 4}}

m2

Ein=m1gH12+m2gH22{E_{in}} = {m_1}g{{{H_1}} \over 2} + {m_2}g{{{H_2}} \over 2}
=ρgA2(H12+H22)=ρgA2(12+1.52)= \rho g{A \over 2}\left( {H_1^2 + H_2^2} \right) = \rho g{A \over 2}\left( {{1^2} + {{1.5}^2}} \right)
Efin=ρgA2(2H2)=ρgA2(2×1.252){E_{fin}} = \rho g{A \over 2}\left( {2{H^2}} \right) = \rho g{A \over 2}\left( {2 \times {{1.25}^2}} \right)
W=ρgA2(3.25−3.125)W = \rho g{A \over 2}\left( {3.25 - 3.125} \right)
=1= 1

J

Q104
A steel wire of length 3.2 m(Ys=2.0×1011 Nm−2)3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \,\mathrm{Nm}^{-2}\right) and a copper wire of length 4.4 m(Yc=1.1×1011 Nm−2)4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \,\mathrm{Nm}^{-2}\right), both of radius 1.4 mm1.4 \mathrm{~mm} are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm1.4 \mathrm{~mm}. The load applied, in Newton, will be: (\quad\left(\right. Given π=227\pi=\dfrac{22}{7})
A 360
B 180
C 1080
D 154
Correct Answer
Option D
Solution
Δls+Δlc=1.4\Delta {l_s} + \Delta {l_c} = 1.4
WlsYs×A+WlcYc×A=1.4×10−3{{W{l_s}} \over {{Y_s} \times A}} + {{W{l_c}} \over {{Y_c} \times A}} = 1.4 \times {10^{ - 3}}
W=1.4×10−3[3.22×(π×1.4×10−3)2+4.41.1×(π×1.4×10−3)2]110+11W = {{1.4 \times {{10}^{ - 3}}} \over {\left[ {{{3.2} \over {2 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}} + {{4.4} \over {1.1 \times {{(\pi \times 1.4 \times {{10}^{ - 3}})}^2}}}} \right]{1 \over {{{10}^{ + 11}}}}}}
W≃154 NW \simeq 154\,N
Q105
The force required to stretch a wire of cross-section 1 cm21 \mathrm{~cm}^{2} to double its length will be : (Given Yong's modulus of the wire =2×1011 N/m2=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2})
A 1×107 N1 \times 10^{7} \mathrm{~N}
B 1.5×107 N1.5 \times 10^{7} \mathrm{~N}
C 2×107 N2 \times 10^{7} \mathrm{~N}
D 2.5×107 N2.5 \times 10^{7} \mathrm{~N}
Correct Answer
Option C
Solution
A=1A = 1

cm2

Y=FlAΔlY = {{Fl} \over {A\Delta l}}
F=YAΔll=2×1011×10−4×llF = {{YA\Delta l} \over l} = {{2 \times {{10}^{11}} \times {{10}^{ - 4}} \times l} \over l}
=2×107= 2 \times {10^7}

N

Q106
A pressure-pump has a horizontal tube of cross sectional area 10 cm210 \mathrm{~cm}^{2} for the outflow of water at a speed of 20 m/s20 \mathrm{~m} / \mathrm{s}. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is : [given: density of water =1000 kg/m3=1000 \mathrm{~kg} / \mathrm{m}^{3}]
A 300 N
B 500 N
C 250 N
D 400 N
Correct Answer
Option D
Solution
Fw=ρAv2{F_w} = \rho A{v^2}
=103×10×10−4×20×20= {10^3} \times 10 \times {10^{ - 4}} \times 20 \times 20
=400 N= 400\,N
Q107
The Young's modulus of a steel wire of length 6 m6 \mathrm{~m} and cross-sectional area 3 mm23 \mathrm{~mm}^{2}, is 2×1011 N/m22 \times 10^{11}~\mathrm{N} / \mathrm{m}^{2}. The wire is suspended from its support on a given planet. A block of mass 4 kg4 \mathrm{~kg} is attached to the free end of the wire. The acceleration due to gravity on the planet is 14\dfrac{1}{4} of its value on the earth. The elongation of wire is (Take gg on the earth =10 m/s2=10 \mathrm{~m} / \mathrm{s}^{2}) :
A 0.1 cm
B 1 cm
C 0.1 mm
D 1 mm
Correct Answer
Option C
Solution

The elongation of the wire can be calculated using the formula for stress and strain.

The stress in the wire is given by:

σ=mgA\sigma = \frac{mg}{A}

where m is the mass of the block (4 kg), g is the acceleration due to gravity on the planet (1/4 of its value on the earth, or 2.5 m/s2), and A is the cross-sectional area of the wire (3 mm2).

The strain in the wire is given by:

ϵ=ΔLL\epsilon = \frac{\Delta L}{L}

where ΔL is the elongation of the wire and L is the original length of the wire (6 m).

Using Hooke's law, which states that stress is proportional to strain, we can find the elongation of the wire:

σ=Yϵ\sigma = Y\epsilon

where Y is the Young's modulus of the wire (2 ×\times 1011 N/m2).

Combining the above equations, we can find the elongation of the wire:

ϵ=σY=mgAY=4×2.53×10−6×2×1011=53×10−6×1011=53×105\epsilon = \frac{\sigma}{Y} = \frac{mg}{A Y} = \frac{4 \times 2.5}{3 \times 10^{-6} \times 2 \times 10^{11}} = \frac{5}{3 \times 10^{-6} \times 10^{11}} = \frac{5}{3 \times 10^{5}}

∴\therefore

ΔLL\frac{\Delta L}{L}
=53×105= \frac{5}{3 \times 10^{5}}

⇒\Rightarrow

ΔL=5×63×105\Delta L = {{5 \times 6} \over {3 \times {{10}^5}}}

=

1104{1 \over {{{10}^4}}}

= 0.1 mm So, the elongation of the wire is 0.1 mm.

Q108
A mercury drop of radius 10−3 m10^{-3}~\mathrm{m} is broken into 125 equal size droplets. Surface tension of mercury is 0.45 Nm−10.45~\mathrm{Nm}^{-1}. The gain in surface energy is :
A 28×10−5 J28\times10^{-5}~\mathrm{J}
B 17.5×10−5 J17.5\times10^{-5}~\mathrm{J}
C 5×10−5 J5\times10^{-5}~\mathrm{J}
D 2.26×10−5 J2.26\times10^{-5}~\mathrm{J}
Correct Answer
Option D
Solution

Initial surface energy =0.45×4π(10−3)2=0.45 \times 4 \pi\left(10^{-3}\right)^2

43π(10−3)3=125×4π3Rnew 3∴10−3=5Rnew ∴Rnew =10−35 m\begin{aligned} & \frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text{new }}^3 \\\\ \therefore & 10^{-3}=5 R_{\text{new }} \\\\ \therefore & R_{\text{new }}=\frac{10^{-3}}{5} \mathrm{~m} \end{aligned}

So, final surface energy =0.45×125×4π(10−35)2=0.45 \times 125 \times 4 \pi\left(\dfrac{10^{-3}}{5}\right)^2 Increase in energy =0.45×4π×(10−3)2[12525−1]=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\dfrac{125}{25}-1\right]

=4×0.45×4π×10−6=2.26×10−5 J\begin{aligned} & =4 \times 0.45 \times 4 \pi \times 10^{-6} \\\\ & =2.26 \times 10^{-5} \mathrm{~J} \end{aligned}
Q109
If 1000 droplets of water of surface tension 0.07 N/m0.07 \mathrm{~N} / \mathrm{m}, having same radius 1 mm1 \mathrm{~mm} each, combine to from a single drop. In the process the released surface energy is :- (Take π=227)\left( {\mathrm{Take}\,\pi = {{22} \over 7}} \right)
A 7.92×10−4 J7 .92 \times 10^{-4} \mathrm{~J}
B 7.92×10−6 J7 .92 \times 10^{-6} \mathrm{~J}
C 8.8×10−5 J8 .8 \times 10^{-5} \mathrm{~J}
D 9.68×10−4 J9 .68 \times 10^{-4} \mathrm{~J}
Correct Answer
Option A
Solution

1000×4π3(1)3=4π3R31000 \times \dfrac{4 \pi}{3}(1)^{3}=\dfrac{4 \pi}{3} \mathrm{R}^{3} R=10 mm\mathrm{R}=10 \mathrm{~mm} T×1000×4π(10−3)2−T×4π(10×10−3)2=ΔE\mathrm{T} \times 1000 \times 4 \pi\left(10^{-3}\right)^{2}-\mathrm{T} \times 4 \pi\left(10 \times 10^{-3}\right)^{2}=\Delta \mathrm{E} ⇒\Rightarrow ΔE=4×π×7×10−2[1000−100]×10−6\Delta \mathrm{E}=4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6} ⇒\Rightarrow ΔE=7.92×10−4 J\Delta \mathrm{E}=7.92 \times 10^{-4} \mathrm{~J}

Q110
A force is applied to a steel wire 'A', rigidly clamped at one end. As a result elongation in the wire is 0.2 mm0.2 \mathrm{~mm}. If same force is applied to another steel wire ' B\mathrm{B} ' of double the length and a diameter 2.42.4 times that of the wire ' A\mathrm{A} ', the elongation in the wire ' B\mathrm{B} ' will be (wires having uniform circular cross sections)
A 6.9×10−2 mm6 .9 \times 10^{-2} \mathrm{~mm}
B 6.06×10−2 mm6.06 \times 10^{-2} \mathrm{~mm}
C 2.77×10−2 mm2.77 \times 10^{-2} \mathrm{~mm}
D 3.0×10−2 mm3.0 \times 10^{-2} \mathrm{~mm}
Correct Answer
Option A
Solution

∵\because

Δl=Fl(4)Yπd2\Delta l = {{Fl(4)} \over {Y\pi {d^2}}}
Δl1Δl2=Δl1d12×d22l2{{\Delta {l_1}} \over {\Delta {l_2}}} = {{\Delta {l_1}} \over {d_1^2}} \times {{d_2^2} \over {{l_2}}}
0.2Δl2=12×(2.4)2{{0.2} \over {\Delta {l_2}}} = {1 \over 2} \times {(2.4)^2}
Δl2=2×0.2(2.4)2\Delta {l_2} = {{2 \times 0.2} \over {{{(2.4)}^2}}}
=6.9×10−2= 6.9 \times {10^{ - 2}}

mm

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