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Question: Raindrops of radius $1 \mathrm{~mm}$ and mass $4 \mathrm{mg}$ are falling with a speed of $30 \mathrm{~m} / \mathrm{s}$ on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head. Solution: From $^{v^{2}}=u^{2}-2 a s$ $a=\frac{v^{2}-u^{2}}{2 s}=\frac{-(30)^{2}}{2 \times 10^{-3}}=-4.5 \times 10^{5} \mathrm{~m} / \mathrm{s}^{2}$ $F=...
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Question: Two blocks $\mathrm{A}$ and $\mathrm{B}$ of mass $\mathrm{mA}$ and $\mathrm{mB}$ respectively are kept in contact on a frictionless table. The experimenter pushes the block $A$ from behind so that the blocks accelerate. If the block $A$ exerts a force $F$ on the block $B$, what is the force exerted by the experimenter on $A$ ? Solution: $F^{\prime}-F=m_{a g}$ $\mathrm{F}=\mathrm{mbg}$ $F^{\prime}=\left(m_{a}+m_{b}\right) g$...
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Question: A particle of mass $50 \mathrm{~g}$ moves on a straight line. The variation of speed with time is shown in figure. Find the force acting on the particle at $t=2,4$ and 6 seconds. Solution: Slope of v-t graph gives acceleration, $\mathrm{t}=25, \mathrm{a}=$ slope of straight line from $\mathrm{t}=0$ to $\mathrm{t}=35=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}^{2}$ $t=45, \quad a=0$ $\left(a s \frac{d v}{d t}=0\right)$ $t=65, \quad a=\frac{-10}{2}=-5 \mathrm{~m} / \mathrm{s}^{2}$ $F(t=25)=0...
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Question: Two blocks of equal mass $m$ are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks. Solution: $\mathrm{F}-\mathrm{T}=\mathrm{ma}$ $\mathrm{F}=2 \mathrm{~T}$ $T=\frac{F}{2}$...
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Question: A block of mass $0.2 \mathrm{~kg}$ is suspended from the ceiling by a light string. A second block of mass $0.3 \mathrm{~kg}$ is suspended from the first block through another string. Find the tensions in the two strings. Take $g=$ $10 \mathrm{~m} / \mathrm{s}^{2}$. Solution: $T 1=0.2 g+T 2$ $T 2=0.3 g=3 N$ $T 1=0.2 g+3=5 N$...
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Question: In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of $5 \times 10^{6} \mathrm{~m} / \mathrm{s}$ in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is $9.1 \times 10^{-31} \mathrm{~kg}$. Solution: $u=0$ (negligible) $v=5 \times 10^{6} \mathrm{~m} / \mathrm{s}$ $v^{2}=u^{2}+2 a s$ $\left(5 \times 10^{6}\right)^{2}=(0)^{2}+2 a\left(1 \times 10^{-2}\right...
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Question: A car moving at $40 \mathrm{~km} / \mathrm{h}$ is to be stopped by applying brakes in the next $4.0 \mathrm{~m}$. If the car weighs 2000 $\mathrm{kg}$, what average force must be applied on it? Solution: $u=40 \mathrm{~km} / \mathrm{hr}=\frac{40 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=\frac{100}{a} \mathrm{~m} / \mathrm{s}$ By law of kinematics, $v^{2}=u^{2}+2 a s$ $0=\left(\frac{100}{a}\right)^{2}+2 a \cdot 4$ $(v=0)$ $a=15.432 \mathrm{~m} / \mathrm{s}^{2}$ $F=m a=2000 \ti...
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Question: A block of mass $2 \mathrm{~kg}$ placed on a long frictionless horizontal table is pulled horizontally by a constant force $F$. It is found to move $10 \mathrm{~m}$ in the first two seconds. Find the magnitude of $F$. Solution: $F=m a$ From law of kinematics, $x-x_{0}=u t+\frac{1}{2} a t^{2}$ $10=\frac{a 2^{2}}{2}$ $(u=0)$ $a=5 m / s^{2}$ $F=m a=2 \times 5=10 \mathrm{~N}$...
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Question: The geostationary orbit of the earth is at a distance of about $36000 \mathrm{~km}$ from the earth's surface. Find the weight of $120 \mathrm{~kg}$ equipment placed in a geostationary satellite. The radius of the earth is $6400 \mathrm{~km}$. Solution: $g=\frac{G M}{R^{2}}$ $g^{\prime}=\frac{G M}{(R+h)^{2}}$ $\frac{g^{\prime}}{g}=\frac{G M /(R+h)^{2}}{G M / R^{2}}=\frac{R^{2}}{(R+h)^{2}}$ $g^{\prime}=\left[\frac{(6400)^{2}}{(36000+6400)^{2}}\right] \times g$ $g^{\prime}=0.0227 \times 9...
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Question: The average separation between the proton and the electron in a hydrogen atom in ground state is $5.3 \times 10^{-11} \mathrm{~m}$. (a) Calculate the Coulomb force between them at this separation (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state? Solution: $F_{C}=\frac{k q_{p} q_{e}}{r^{2}}=\frac{9 \times 10^{9} \times\left(1.6 \tim...
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Question: Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons. Solution: $63 \times R$ SAQ - 10 Find the ratio. $F_{e}=\frac{k q_{p} q_{p}}{r^{2}}$ $F_{e}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{r^{2}}$ $F_{G}=\frac{6.67 \times 10^{11} \times\left(1.732 \times 10^{-27}\right)^{2}}{r^{2}}$ Ratio is $\frac{F_{e}}{F_{G}}=\frac{\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{r^{2}}}{\frac{6.67 \times ...
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Question: The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data: The universal constant of gravitation $G=$ $6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^{2} / \mathrm{kg}$, mass of the moon $=7.36 \times 10^{22} \mathrm{~kg}$, mass of the earth $=6 \times 10^{24} \mathrm{~kg}$ and the distance between the earth and the moon $=3.8 \times 10^{5} \mathrm{~km}$. Solution: $W_{M}=F=\frac{G M_{e} M_{M}}{R^{2}}$ $=...
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Question: Two charged particles placed at a separation of $20 \mathrm{~cm}$ exert $20 \mathrm{~N}$ of Coulomb force on each other. What will be the force if the separation is increased to $25 \mathrm{~cm}$ ? Solution: $F=\frac{k q_{1} q_{2}}{r^{2}}$ $F \propto \frac{1}{r^{2}}$ $\frac{F}{F^{\prime}}=\frac{r^{\prime 2}}{r^{2}}$ $F^{\prime}=\frac{20 \mathrm{x}(0.20)^{2}}{(0.25)^{2}}$ $F^{\prime}=12.8 \mathrm{~N}$...
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Question: A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is $6400 \mathrm{~km}$.) Solution: $F_{h}=\frac{1}{2} F_{e}$ $\frac{G M m}{(R+h)^{2}}=\frac{1}{2} \frac{G M m}{R^{2}}$ $(R+h)^{2}=2 R^{2}$ $h^{2}+2 R h-R^{2}=0$ Solving equation, $h=\frac{-2 R \pm \sqrt{4 R^{2}+4 R^{2}}}{2}$ $h=-R \pm \sqrt{2} R$ $h=R(-1 \pm \s...
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Question: A body builder exerts a force of $150 \mathrm{~N}$ against a bullworker and compresses it by $20 \mathrm{~cm}$. Calculate the spring constant of the spring in the bullworker. Solution: Net Force $F=k x$ $150=k \times 0.2$ $\mathrm{k}=750 \mathrm{~N} / \mathrm{m}$...
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Question: A monkey is sitting on a tree limb. The limb exerts a normal force of $48 \mathrm{~N}$ and a frictional force of 20 $\mathrm{N}$. Find the magnitude of the total force exerted by the limb on the monkey. Solution: $F_{N}=48 \mathrm{~N}$ and $F_{F}=20 \mathrm{~N}$ $F_{R}=\sqrt{F_{N}^{2}+F_{F}^{2}}$ $F_{R}=\sqrt{48^{2}+20^{2}}$ $F_{R}=52 \mathrm{~N}$...
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Question: Two spherical bodies, each of mass $50 \mathrm{~kg}$, are placed at a separation of $20 \mathrm{~cm}$. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body. Solution: $F_{1}=\frac{G m_{1} m_{\Omega}}{r^{2}}$ $F_{1}=\frac{6.67 \times 10^{-11} \times 2500}{(0.2)^{2}}$ Coulomb's Force $F_{2}=\frac{k q_{1} q_{2}}{r^{2}}$ $F_{2}=\frac{9 \times 10^{9}...
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Question: At what distance should two charges, each equal to $1 \mathrm{C}$, be placed so that the force between them equals your weight? Solution: $F=W$ $\frac{k q_{1} q_{2}}{r^{2}}=m g$ $\frac{9 \times 10^{9} \times 1 \times 1}{r^{2}}=m \times 10$ $r=\frac{3 \times 10^{4}}{\sqrt{m}}$ meters If $m=49 \mathrm{~kg}$ $r=4285.7 \mathrm{~m}$...
Read More →Calculate the force with which you attract the earth.
Question: Calculate the force with which you attract the earth. Solution: Force acting on us $=\mathrm{mg}$ Let $m=60 \mathrm{~kg}$ Then, $F=W=m \times g=60 \times 10$ $F=600 \mathrm{~N}$....
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Question: The gravitational force acting on a particle of $1 \mathrm{~g}$ due to a similar particle is equal to $6.67 \times 10^{-11} \mathrm{~N}$. Calculate the separation between the particles. Solution: $F=\frac{G m_{1} m_{2}}{r^{2}}$ $6.67 \times 10^{-17}=\frac{6.67 \times 10-11 \times \frac{1}{1000} \times \frac{1}{1000}}{x^{2}}$ $r^{2}=\frac{6.67 \times 10-17}{6.67 \times 10-17}$ $r=1 \mathrm{~m} .$...
Read More →Action and reaction
Question: Action and reactionact on two different objectshave equal magnitudehave opposite directionshave resultant zeroCorrect Option: 1, 2, 3, 4 Solution: According to the third law of motion, every action has equal and opposite reaction. Thus, action and reaction act on different objects, have equal magnitude, have opposite direction and have resultant zero. (1), (2), (3), (4)...
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Question: The two ends of a spring are displaced along the length of the spring. All displacements have equal magnitudes. In which case or cases the tension or compression in the spring will have a maximum magnitude?the right end is displaced towards right and the left end towards leftboth ends are displaced towards rightboth ends are displaced towards leftthe right end is displaced towards left and the left end towards right.Correct Option: 1, 4 Solution: Maximum magnitude in tension is when th...
Read More →Which of the following systems may be adequately described by classical physics?
Question: Which of the following systems may be adequately described by classical physics?motion of a cricket ballmotion of a dust particlea hydrogen atoma neutron changing to a protonCorrect Option: 1, 2 Solution: Classical physics is used generally for heavenly bodies and do not work for smaller size particle. Thus motion od cricket ball, motion of dust particles is adequately described by classical physics. (1), (2)...
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Question: If all matter were made of electrically neutral particles such as neutronsthere would be no force of frictionthere would be no tension in the stringit would not be possible to sit on a chairthe earth could not move around the sunCorrect Option: 1, 2, 3 Solution: If all matter were made of electrically neutral particles, then there would be no friction, no tension in the string and also it would not be possible to sit on a chair. But earth could move around the sun as it is due to gravi...
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Question: Mark the correct statements:The nuclear force between two protons is always greater than the electromagnetic force between them.The electromagnetic force between two protons is always greater than the gravitational force between them.The gravitational force between two protons may be greater than the nuclear force between them.Electromagnetic force between two protons may be greater than the nuclear force acting between themCorrect Option: 1, 2, 3 Solution: Electromagnetic force is gre...
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