Q: A uniform chain of length \$L\$ and mass \$M\$ is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If \$g\$ is acceleration due to gravity, work required to pull the hanging part on to the table is:

\$\textbf{Step 1: Figure and Initial condition}\$ \$\textbf{[Refer Fig. 1 & 2]}\$\$\textbf{Step 2: Work done due to small element of length dx}\$ Mass of \$dx\$ length \$=dx \times \dfrac{m}{L}\$ Work done \$dW = dmgh\$ \$dW = dx \times \dfrac{m}{L} gh\$ \${h=x}\$ \$dW = \dfrac{m}{L} gxdx\$\$\textbf{Step 3: Intigration and Net workdone}\$ \$\displaystyle \int_0^W dW = \int_0^{L/3} \dfrac{m}{L} gx dx\$ (length of hanging part \$=L/3\$) \$W = \dfrac{m}{L} g \left[\dfrac{x^2}{2}\right]_0^{L/3}\$ \$W = \dfrac{m}{L} g \dfrac{L^2}{18} = \dfrac{mgL}{18} \ J\$Option \$(D)\$ is correct.\$\textbf{Alternative:}\$ Mass of hanging part \$(m') = \dfrac{m}{3}\$Distance moved by C.O.M. of hanging part \$= \dfrac{L}{6}\$ \$\textbf{[Refer Fig. 1]}\$So, Workdone \$= m'gh\$ \$= \left(\dfrac{m}{3}\right) \times g \times \dfrac{L}{6}\$ \$[{h = \dfrac{L}{6}}]\$ \${W = \dfrac{mgL}{18} \ J}\$Common mistakes:\$\bullet\$ Use the mass of hanging part only\$\bullet\$ Distance moved will be taken from the center of mass.

Q: If the kinetic energy of a body is increased by 300%, its momentum will increase by

Kinetic energy \$k =\frac {P^2}{2m}\$where, P is momentum and m is mass. \$k_1=k, k_2=k+300\$ % of \$k=4k\$\$\Rightarrow \frac {P_2}{P_1}=\sqrt {\frac{k_2}{k_1}}=\sqrt{\frac{4k}{k}}=2\$\$\Rightarrow P_2=P_1 + 100\$% pf \$P_1\$Hence, momentum will increase by 100%.

Q: State and prove law of conservation of energy in case of freely falling body?

Law of conservation of energy states that\$:-\$Energy can neither be created nor be destroyed\$.\$Energy transforms from one form to another form and\$,\$ potential energy\$+\$ kinetic energy\$=\$ constantProof\$:-\$At point \$A\$\$K.E = 0\,m/s\$\$P.E = mgh\$\$P.E = mgh\$At point \$B,\$\$K.E.,\$\$2gh = {V^2} - {O^2}\$\${V^2} = 2gx\$\$K.E = \frac{1}{2}m{V^2}\$\$K.E = \frac{1}{2}m.2gx\$\$K.E = mgx\$-----------------------------\$(1)\$\$P.E = mgh\$\$P.E = m.g.\left( {h - x} \right)\$\$P.E = mgh - mgx\$---------------------\$(2)\$from\$(1)\$ and \$(2)\$\$K.E+P.E\$ constant\$mgx+mgh-mgx=mgh\$\$mgh=mgh\$At point \$C,\$ \$P.E=0\$\$2gh = {v^2} - {u^2}\$ \$(\$ initial velocity \$=0\$ m/s\$)\$\$2gh = {v^2}\$\$K.E = \frac{1}{2}.m.v\$ \$K.E = \frac{1}{2}.m.2gh\$\$K.E = mgh\$Thus\$,\$ in all the points the energy is same\$.\$

Q: Find the components of \$\vec{a} = 2\hat{i} + 3\hat{j}\$ along the directions of vectors \$\hat{i} + \hat{j}\$ and \$\hat{i} - \hat{j}\$:

\$\textbf{Step 1: Calculation of component of}\,\, \vec{a}\,\, \text{along}\, \hat{i} + \hat{j} \$Let \$\vec{b} = \hat{i} + \hat{j} \$The unit vector \$\hat{b}\$\$\hat{b} = \dfrac{\hat{i} + \hat{j}}{| \hat{i} + \hat{j}|} = \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \$ Component of vector \$\vec{a} \$ along \$\vec{b}\$ vector \$ = a \cos \theta \$ \$ = a \dfrac{\vec{a} . \vec{b}}{|ab|} = \dfrac{\vec{a} . \vec{b}}{b} = \vec{a} . \hat{b} \$ To get the vector form, \$(\vec{a} . \hat{b}) \hat{b} \$ \$ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \$ \$= \left [\dfrac{1}{\sqrt{2}}(2 + 3) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \$ \$ = \dfrac{5}{2} (\hat{i} + \hat{j}) \$ \$\textbf{Step 2: Component of}\ \vec{a}\,\, \textbf{along}\,\, \hat{i} - \hat{j} \$Let \$\vec{c} = \hat{i} - \hat{j} \$ Unit vector, \$\hat{c} = \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \$ Component of vector \$\vec{a}\$ along \$\vec{c} \$ vector \$ = (\vec{a} . \hat{c}) \hat{c} \$ \$ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \$ \$ = \left [-\dfrac{1}{\sqrt{2}} \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \$ \$ = \dfrac{-1}{2} \hat{i} + \dfrac{1}{2} \hat{j} \$

Q: A pump on the ground floor of a building can pump up water to fill a tank of volume \$30m^{3}\$ in \$15\$ min. If the tank is \$40\ m\$ above the ground, and the efficiency of the pump is \$30\$%, how much electric power is consumed by the pump?(Take \$g = 10\ ms^{-2})\$.

Here, Volume of water \$= 30\ m^{3}\$\$t = 15\ min = 15\times 60\ s = 900\ s; Height, h = 40\ m\$Efficiency, \$\eta = 30\$%Density of water \$= 10^{3} kg\ m^{-3}\$\$\therefore\$ Mass of water pumped \$= Volume \times Density\$\$= (30\ m^{3})(10^{3}\ kg\ m^{-3}) = 3\times 10^{4} kg\$\$P_{output} = \dfrac {W}{t} = \dfrac {mgh}{t} = \dfrac {(3\times10^{4}kg)(10\ ms^{-2})(40\ m)}{900\ s}\$\$= \dfrac {4}{3} \times 10^{4} W\$Efficiency, \$\eta = \dfrac {P_{output}}{P_{input}}\$\$P_{input} = \dfrac {P_{output}}{\eta} = \dfrac {4\times 10^{4}}{3\times \dfrac {30}{100}} = \dfrac {4}{9} \times 10^{5}\$\$= 44.4\times 10^{3}W = 44.4\ kW\$.

Q: If the linear momentum of a body is increased by 50 %, the Kinetic energy will be increased by

If the initial momentum is \$p_0\$, the initial kinetic energy is \$K_0=\dfrac{p_0^2}{2m}\$When momentum is increased by 50 %, the momentum becomes \$p=p_0+(p_0\times \dfrac{50}{100})=(3/2)p_0\$ Now kinetic energy, \$K=\dfrac{p^2}{2m}=(9/4)\times (p_0^2/2m)=(9/4)K_0\$Thus the percentage increased in kinetic energy \$=\dfrac{K-K_0}{K_0}\times 100=[(9/4)-1]100=125\$ %

Q: Which one of the following is not the unit of energy?

\$\textbf{Hint:}\$ Use the relation between energy and power.\$\textbf{Explanation :}\$If Watt is unit of power then kilowatt is also the unit of power.Work done = Force ×× Distance = Newton - metre, Therefore it is the unit of Work or Energy.Energy = Power ×× time = kilowatt - hour, in fact it is the commercial unit of Energy.

Q: Which one of the following statements is not true?

We know that force is equal to rate of change of momentum. \$\implies F = \dfrac{dp}{dt}\$\$\implies F \propto dp\$From the given equation we can easily observed thatlarger the change in momentum in a given time, larger will be the force.So , option (D) is incorrect statement.

Question 1

A uniform chain of length $L$ and mass $M$ is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If $g$ is acceleration due to gravity, work required to pull the hanging part on to the table is:

Accepted Answer

$\textbf{Step 1: Figure and Initial condition}$     $\textbf{[Refer Fig. 1 & 2]}$$\textbf{Step 2: Work done due to small element of length dx}$     Mass of $dx$ length $=dx \times \dfrac{m}{L}$     Work done $dW = dmgh$                $dW = dx \times \dfrac{m}{L} gh$         ${h=x}$                $dW = \dfrac{m}{L} gxdx$$\textbf{Step 3: Intigration and Net workdone}$                $\displaystyle \int_0^W dW = \int_0^{L/3} \dfrac{m}{L} gx dx$     (length of hanging part $=L/3$)                $W = \dfrac{m}{L} g \left[\dfrac{x^2}{2}\right]_0^{L/3}$                $W = \dfrac{m}{L} g \dfrac{L^2}{18} = \dfrac{mgL}{18} \ J$Option $(D)$ is correct.$\textbf{Alternative:}$ Mass of hanging part $(m') = \dfrac{m}{3}$Distance moved by C.O.M. of hanging part $= \dfrac{L}{6}$       $\textbf{[Refer Fig. 1]}$So, Workdone $= m'gh$                   $= \left(\dfrac{m}{3}\right) \times g \times \dfrac{L}{6}$                             $[{h = \dfrac{L}{6}}]$            ${W = \dfrac{mgL}{18} \ J}$Common mistakes:$\bullet$ Use the mass of hanging part only$\bullet$ Distance moved will be taken from the center of mass.

Question 2

If the kinetic energy of a body is increased by 300%, its momentum will increase by

Accepted Answer

Kinetic energy $k =\frac {P^2}{2m}$where, P is momentum and m is mass. $k_1=k, k_2=k+300$ % of $k=4k$$\Rightarrow \frac {P_2}{P_1}=\sqrt {\frac{k_2}{k_1}}=\sqrt{\frac{4k}{k}}=2$$\Rightarrow P_2=P_1 + 100$% pf $P_1$Hence, momentum will increase by 100%.

Question 3

State and prove law of conservation of energy in case of freely falling body?

Accepted Answer

Law of conservation of energy states that$:-$Energy can neither be created nor be destroyed$.$Energy transforms from one form to another form and$,$ potential energy$+$ kinetic energy$=$ constantProof$:-$At point $A$$K.E = 0\,m/s$$P.E = mgh$$P.E = mgh$At point $B,$$K.E.,$$2gh = {V^2} - {O^2}$${V^2} = 2gx$$K.E = \frac{1}{2}m{V^2}$$K.E = \frac{1}{2}m.2gx$$K.E = mgx$-----------------------------$(1)$$P.E = mgh$$P.E = m.g.\left( {h - x} \right)$$P.E = mgh - mgx$---------------------$(2)$from$(1)$ and $(2)$$K.E+P.E$ constant$mgx+mgh-mgx=mgh$$mgh=mgh$At point $C,$ $P.E=0$$2gh = {v^2} - {u^2}$ $($ initial velocity $=0$ m/s$)$$2gh = {v^2}$$K.E = \frac{1}{2}.m.v$ $K.E = \frac{1}{2}.m.2gh$$K.E = mgh$Thus$,$ in all the points the energy is same$.$

Question 4

Find the components of $\vec{a} = 2\hat{i} + 3\hat{j}$ along the directions of vectors $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$:

Accepted Answer

$\textbf{Step 1: Calculation of component of}\,\, \vec{a}\,\, \text{along}\, \hat{i} + \hat{j}  $Let $\vec{b} = \hat{i} + \hat{j} $The unit vector $\hat{b}$$\hat{b} = \dfrac{\hat{i} + \hat{j}}{| \hat{i} + \hat{j}|} = \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) $ Component of vector $\vec{a} $ along $\vec{b}$ vector $ = a \cos \theta $ $ = a \dfrac{\vec{a} . \vec{b}}{|ab|} = \dfrac{\vec{a} . \vec{b}}{b} = \vec{a} . \hat{b} $ To get the vector form,   $(\vec{a} . \hat{b}) \hat{b} $ $ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) $   $= \left [\dfrac{1}{\sqrt{2}}(2 + 3) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} + \hat{j}) $ $ = \dfrac{5}{2} (\hat{i} + \hat{j}) $ $\textbf{Step 2: Component of}\ \vec{a}\,\, \textbf{along}\,\, \hat{i} - \hat{j} $Let $\vec{c} = \hat{i} - \hat{j} $ Unit vector, $\hat{c} = \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) $ Component of vector $\vec{a}$ along $\vec{c} $ vector $ = (\vec{a} . \hat{c}) \hat{c} $ $ = \left [ (2 \hat{i} + 3 \hat{j}) . \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) $     $ = \left [-\dfrac{1}{\sqrt{2}} \right ] \dfrac{1}{\sqrt{2}} (\hat{i} - \hat{j}) $ $ = \dfrac{-1}{2} \hat{i} + \dfrac{1}{2} \hat{j} $

Question 5

A pump on the ground floor of a building can pump up water to fill a tank of volume $30m^{3}$ in $15$ min. If the tank is $40\ m$ above the ground, and the efficiency of the pump is $30$%, how much electric power is consumed by the pump?(Take $g = 10\ ms^{-2})$.

Accepted Answer

Here, Volume of water $= 30\ m^{3}$$t = 15\ min = 15\times 60\ s = 900\ s; Height, h = 40\ m$Efficiency, $\eta = 30$%Density of water $= 10^{3} kg\ m^{-3}$$\therefore$ Mass of water pumped $= Volume \times Density$$= (30\ m^{3})(10^{3}\ kg\ m^{-3}) = 3\times 10^{4} kg$$P_{output} = \dfrac {W}{t} = \dfrac {mgh}{t} = \dfrac {(3\times10^{4}kg)(10\ ms^{-2})(40\ m)}{900\ s}$$= \dfrac {4}{3} \times 10^{4} W$Efficiency, $\eta = \dfrac {P_{output}}{P_{input}}$$P_{input} = \dfrac {P_{output}}{\eta} = \dfrac {4\times 10^{4}}{3\times \dfrac {30}{100}} = \dfrac {4}{9} \times 10^{5}$$= 44.4\times 10^{3}W = 44.4\ kW$.

Question 6

If the linear momentum of a body is increased by 50 %, the Kinetic energy will be increased by

Accepted Answer

If the initial momentum is $p_0$, the  initial kinetic energy is $K_0=\dfrac{p_0^2}{2m}$When momentum is increased by 50 %, the momentum becomes $p=p_0+(p_0\times \dfrac{50}{100})=(3/2)p_0$ Now kinetic energy, $K=\dfrac{p^2}{2m}=(9/4)\times (p_0^2/2m)=(9/4)K_0$Thus the percentage increased in kinetic energy $=\dfrac{K-K_0}{K_0}\times 100=[(9/4)-1]100=125$ %

Question 7

Which one of the following is not the unit of energy?

Accepted Answer

$\textbf{Hint:}$ Use the relation between energy and power.$\textbf{Explanation :}$If Watt is unit of power then kilowatt is also the unit of power.Work done = Force &#215;&#215; Distance = Newton - metre, Therefore it is the unit of Work or Energy.Energy = Power &#215;&#215; time = kilowatt - hour, in fact it is the commercial unit of Energy.

Question 8

Which one of the following statements is not true?

Accepted Answer

We know that force is equal to rate of change of momentum. $\implies F = \dfrac{dp}{dt}$$\implies  F \propto dp$From the given equation we can easily observed thatlarger the change in momentum in a given time, larger will be the force.So , option (D) is incorrect statement.

Question 9

State work-energy theorem. Prove it for a variable force.

Accepted Answer

Work energy theorem -It states that the work done by the net force acting on a body is equal to the changed produced in kinetic energy of the body. Let F be the variable force $\therefore$ Work done by this variable force , $W=\int_{x_i}^{x_f} F\cdot dx$where $x_i$ is the initial position and $x_f$ is final position.Also Kinetic energy of an object , $K=\dfrac 12 mv^2$ $\implies \dfrac{dK}{dt}=mv \dfrac{dv}{dt}$$\implies \dfrac{dK}{dt}= ma\dfrac{dx}{dt}$$\implies \dfrac{dK}{dt}F\dfrac{dx}{dt}$$\implies dK=Fdx$$\implies \int _{K_i}^{K_f}dK=\int_{x_i}^{x_f} F\cdot dx$$\implies \Delta K= W$Where $\Delta K $ is the change in kinetic energy

Question 10

State and derive work energy theorem.

Accepted Answer

Work energy theorem states that the change in kinetic energy of an object is equal to the net-work done on it by the net force.Let us suppose that a body is initially at rest and a force. $\vec F $is applied on the body to displace it through$ d \vec  s$along the direction of the force. Then, a small amount of work done is given bydw = $\vec F . d \vec s $ = FdsAlso, according to Newton's second law of motion, we haveF = ma where a is acceleration produced (in the direction of force) on applying the force. Therefore, dw = Mada = $M \dfrac{dv}{dt} ds$Now, work done by the force in order to increase its velocity from u (initial velocity) to v (final velocity) is given byW = $\int_u^v Mvdv $ = $M \int_u^v vdv$     = $M|\dfrac{v^2}{2}|_u^v$W = $\dfrac{1}{2} Mv^2 - \frac{1}{2}Mu^2$Hence, work done on a body by a force is equal to the change in its kinetic energy.