Physics 2007 Jamb Examination
Physics 2007 Jamb Examination
Question 41
In a large telecommunications auditorium, perforated absorbent materials are used to line the ceiling so as to
Perforated absorbent materials absorb incident sound energy, reducing reflected sound and therefore decreasing reverberation time. This improves speech intelligibility in large halls. The purpose is to reduce reverberation, not increase it or change the physical height.
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Question 42
A reservoir 500 m deep is filled with a fluid of density 850 kg·m-3. If the atmospheric pressure is $1.05\times10^{5}\ \text{N·m}^{-2}$, the pressure at the bottom of the reservoir is (take $g=10\ \text{m·s}^{-2}$).
Fluid pressure at depth is $p=p_{atm}+\rho g h$. \[p = 1.05\times10^{5} + (850)(10)(500)\]\[p = 1.05\times10^{5} + 4.25\times10^{6} = 4.355\times10^{6}\ \text{N·m}^{-2}.\]Â
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Question 43
A piece of iron weighs 250 N in air and 200 N in a liquid of density $1000\ \text{kg·m}^{-3}$. The volume of the iron is?
Buoyant force = weight in air − apparent weight in liquid = $250-200=50\ \text{N}$.
This buoyant force equals weight of displaced liquid: $B=\rho g V$.
Solve for $V$. \[V=\dfrac{B}{\rho g}=\dfrac{50}{1000\times10}=\dfrac{50}{10000}=5.0\times10^{-3}\ \text{m}^3.\]Â
This buoyant force equals weight of displaced liquid: $B=\rho g V$.
Solve for $V$. \[V=\dfrac{B}{\rho g}=\dfrac{50}{1000\times10}=\dfrac{50}{10000}=5.0\times10^{-3}\ \text{m}^3.\]Â
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Question 44
A clinical thermometer is different from other mercury-in-glass thermometers owing to
Clinical thermometers have a constriction (narrow section) in the capillary that prevents the mercury column from falling back immediately; this allows reading the maximum temperature reached. This feature distinguishes them from ordinary thermometers.
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Question 45
I. Use a liquid with a high melting point.
II. Use a liquid of high volume expansivity.
III. Use a capillary tube of large diameter.
Which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?
II. Use a liquid of high volume expansivity.
III. Use a capillary tube of large diameter.
Which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?
Sensitivity is increased by choosing a liquid with large volumetric expansivity (so a small temperature change produces a larger column movement). A narrow capillary (not large) also increases sensitivity, so statement II alone is correct.
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Question 46
A 2 kg mass of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. What is the increase in temperature of the water? (specific heat capacity of water = $4200\ \text{J·kg}^{-1}\text{K}^{-1}$)
Electrical power: $P=VI=200\times3.5=700\ \text{W}$. Energy supplied in $t=120\ \text{s}$: $Q=Pt=700\times120=84{,}000\ \text{J}$. Temperature rise:
\[\Delta T=\dfrac{Q}{mc}=\dfrac{84{,}000}{2\times4200}=\dfrac{84{,}000}{8{,}400}=10\ ^\circ\text{C}.\]
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Question 47
The unit of moment of a couple can be expressed in?
Moment of a couple (torque) has dimensions force × distance, so the SI unit is newton-metre ($\text{N·m}$).
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Question 48
A gramophone record takes 5 s to reach its constant angular velocity of $4\pi\ \text{rad·s}^{-1}$ from rest. Find its constant angular acceleration.
With initial angular velocity $\omega_0=0$, final $\omega=4\pi$, time $t=5\ \text{s}$, constant angular acceleration $\alpha=\Delta\omega/\Delta t$.
\[\alpha=\dfrac{4\pi-0}{5}=\dfrac{4\pi}{5}=0.8\pi\ \text{rad·s}^{-2}.\]
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Question 49
A car of mass 1500 kg goes round a circular curve of radius 50 m at a speed of $40\ \text{m·s}^{-1}$. The magnitude of the centripetal force on the car is?
Centripetal force $F=mv^2/r$.
\[F=\dfrac{1500\times40^2}{50}=\dfrac{1500\times1600}{50}=1500\times32=48{,}000\ \text{N}=4.8\times10^{4}\ \text{N}.\]
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Question 50
Two spheres of masses 5.0 kg and 10.0 kg are 0.3 m apart. Calculate the force of attraction between them. ($G=6.67\times10^{-11}\ \text{N·m}^2\text{·kg}^{-2}$)
Newton's law of gravitation: \[F=G\dfrac{m_1m_2}{r^2}=6.67\times10^{-11}\times\dfrac{5\times10}{0.3^2}.\] Compute numerator: $6.67\times10^{-11}\times50=3.335\times10^{-9}$.
Denominator $0.09$. \[F=\dfrac{3.335\times10^{-9}}{0.09}\approx3.71\times10^{-8}\ \text{N}.\]
Denominator $0.09$. \[F=\dfrac{3.335\times10^{-9}}{0.09}\approx3.71\times10^{-8}\ \text{N}.\]