Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Nuclear Physics
A nuclear transformation is denoted by $X(n, \alpha)_{3} L i^{7}$. Which of the following is the nucleus of element $X$ ?
Options: (a) ${ }_{5} \mathrm{~B}^{9}$
(b) ${ }_{4} \mathrm{Be}^{11}$
(c) ${ }_{6} \mathrm{C}^{12}$
(d) ${ }_{5} \mathrm{~B}^{10}$
Answer: (d) $5^{10}$
Solution: The nuclear transformation is given by
${ }_{2} X^{A}+{ }_{o n}{ }^{1} \rightarrow{ }_{2} \mathrm{He}^{4}+{ }_{3} \mathrm{Li}^{7}$
According to conservation of mass number
$A+1=4+7$ or $A=10$
According to conservation of charge number
$Z+0 \rightarrow 2+3$ or $Z=5$
So the nucleus of the element be ${ }_{5} \mathrm{~B}^{10}$.
Topic: Nuclear Physics
Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At $t=0$ it was 1600 counts per second and at $t=8$ seconds it was 100 counts per second. The count rate observed, as counts per second, at $t=6$ seconds is close to
Options: (a) 200
(b) 360
(c) 150
(d) 400
Answer: (a) 200
Solution: According to law of radioactivity, the count rate at $\mathrm{t}=8$ seconds is
$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$
$\mathrm{dN} / \mathrm{dt}=\lambda N=\lambda N_{0} \mathrm{e}^{-\lambda t}$
$1600=\lambda N_{0} e^{0}=\lambda N_{0} \Rightarrow 100=\lambda N_{0} e^{-8 \lambda}=1600 e^{-8 \lambda}$
At $t=6 \mathrm{sec}$
$d N / d t=\lambda N_{0} e^{-6 \lambda}=1600 \times\left(e^{-2 \lambda}\right)^{3}=1600 \times(18)=200$
Topic: Thermodynamics
Heat cannot by itself flow from a body at a lower temperature to a body at a higher temperature is a statement or consequence of
Options: (a) The second law of thermodynamics
(b) conservation of momentum
(c) conservation of mass
(d) The first law of thermodynamics
Answer: Second law of thermodynamics.
Topic: Thermodynamics
Which of the following is incorrect regarding the first law of thermodynamics?
Options: (a) It introduces the concept of internal energy
(b) It introduces the concept of entropy
(c) It is not applicable to any cyclic process
(d) It is a restatement of the principle of conservation of energy
Answer: Statements (b) and (c) are incorrect regarding the first law of thermodynamics.
Topic: Thermodynamics
Which of the following statements is correct for any thermodynamic system?
Options: (a) The internal energy changes in all processes
(b) Internal energy and entropy are state functions
(c) The change in entropy can never be zero
(d) The work done in an adiabatic process is always zero
Answer: Internal energy and entropy are state functions
Topic: Thermodynamics
Which of the following parameters does not characterize the thermodynamic state of matter?
Options: (a) temperature
(b) pressure
(c) work
(d) volume
Answer: The work does not characterize the thermodynamic state of matter
Topic: Thermodynamics
Even a Carnot engine cannot give 100 % efficiency because we cannot
Options: (a) prevent radiation
(b) find ideal sources
(c) reach absolute zero temperature
(d) eliminate friction.
Answer: We cannot reach absolute zero temperature
Topic: Thermodynamics
Which statement is incorrect?
Options: (a) All reversible cycles have the same efficiency
(b) The reversible cycle has more efficiency than an irreversible one
(c) Carnot cycle is a reversible one
(d) Carnot cycle has the maximum efficiency in all cycles
Answer: All reversible cycles do not have the same efficiency.
Topic: Thermodynamics
A Carnot engine operating between temperatures T1 and T2 has efficiency 1 / 6 When T2 is lowered by 62 K, its efficiency increases to 1 / 3. Then T1 and T2 are, respectively
Options: (a) 372 K and 310 K
(b) 372 K and 330 K
(c) 330 K and 268 K
(d) 310 K and 248 K
Answer: 372 K and 310 K
Solution: The efficiency of Carnot engine, η=1−(T2/T1) η=1/6 T2/T1=5/6 T1=6 T2/5 As per the question, when T2 is lowered by 62 K, then its efficiency becomes 1/3 1/3=[1−(T2−62/T1)] [T2−62/T1]=1−(1/3) (Using equa (1)) 5(T2−62)/6 T2=2/3 5 T2−310=4 T2 ⇒ T2=310 K From equation (1) T1=(6×310)/5=372 K
Topic: Thermodynamics
100 g of water is heated from 30∘ C to 50∘ C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 Jkg−1 K−1 )
Options: (a) 4.2 kJ
(b) 8.4 kJ
(c) 84 kJ
(d) 2.1 kJ
Answer: 8.4 kJ
Solution: Δ Q=ms Δ T Here m=100 g=100 ×10−3 Kg S=4184 J kg−1 K−1 and Δ T=(50−30)=20∘ C Δ Q=100 ×10−3 ×4184 ×20=8.4 ×103 J Δ Q=Δ U+Δ W Change in internal energy Δ U=Δ Q=8.4 ×103 J=8.4 kJ
Topic: Thermodynamics
200 g water is heated from 40∘ C to 60∘ C. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water =4184 J / kg / K )
Options: (a) 167.4 kJ
(b) 8.4 kJ
(c) 4.2 kJ
(d) 16.7 kJ
Answer: 16.7 kJ
Solution: For isobaric process, Δ U=Q=ms Δ T Here, m=200 g=0.2 Kg, s=4184 J / Kg / K Δ T=60∘ C−40∘ C=20∘ C Δ U=(0.2)(4184)(20)=16736 J=16.7 kJ
Topic: Thermodynamics
The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7∘ C. The gas is (R=8.3 J mol−1 K−1)
Options: (a) monoatomic
(b) diatomic
(c) triatomic
(d) a mixture of monoatomic and diatomic
Answer: diatomic
Solution: According to the first law of thermodynamics Δ Q=Δ U+Δ W For an adiabatic process, Δ Q=0 0=Δ U+Δ W Δ U=−Δ W nC v Δ T=−Δ W C v=−Δ W /n Δ T=−(−146)×103 /[(1 ×103) ×7]=20.8 Jmol−1 K−1 For diatomic gas, C v=(5/2) R=(5/2) ×8.3=20.8 Jmol−1 K−1 Hence, the gas is diatomic
Topic: Thermodynamics
From the following statements, concerning ideal gas at any given temperature T, select the correct.
Options: (a)The coefficient of volume expansion at constant pressure is the same for all ideal gases
(b)The average translational kinetic energy per molecule of oxygen gas is 3 kT, k being Boltzmann constant
(c)The mean-free path of molecules increases with an increase in the pressure
(d)In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different
Answer: The coefficient of volume expansion at constant pressure is the same for all ideal gases
Solution: Y=d V /(V 0 ×d T) at a constant temperature Y=1/V 0(d V/d T) p since P V=R T PdV=RdT or (dV/dT)=R/P 0 Therefore, ϕ=(1/V 0)(R/P 0)=R/R T 0 γ=1/T 0 γ=1/273
Topic: Thermodynamics
Calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 ∘ C and it is defined under which of the following conditions?
Options: (a) From 14.5∘ C to 15.5∘ C at 760 mm of Hg
(b) From 98.5∘ C to 99.5∘ C at 760 mm of Hg
(c) From 13.5∘ C to 14.5∘ C at 76 mm of Hg
(d) From 3.5∘ C to 4.5∘ C at 76 mm of Hg
Answer: From 14.5∘ C to 15.5∘ C at 760 mm of Hg
Solution: 1 calorie is the amount of heat required to raise the temperature of 1 gm of water from 14.5∘ C to 15.5 ∘ C at 760 mm of Hg
Topic: Thermodynamics
The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N 2 (molar mass 28) molecules in eV at the same temperature is
Options: (a) 0.0015
(b) 0.003
(c) 0.048
(d) 0.768
Answer: 0.048
Solution: Average Kinetic Energy =(3/2) KT It depends on temperature and does not depend on molar mass For both the gases, average translational kinetic energy will be the same ie., 0.048 eV
Topic: Thermodynamics
One mole of a monoatomic gas is heated at a constant pressure of 1 atmosphere from 0 K to 100 K. If the gas constant R=8.32 J / mol K, the change in internal energy of the gas is approximately [1998]
Options: (a) 2.3 J
(b) 46 J
(c) 8.67 ×103 J
(d) 1.25 ×103 J
Answer: 1.25 ×103 J
Solution: Δ U=C v dT=(3 R/2) Δ T Δ U=(3/2) ×(8.3) ×(100)=1.25 ×103 J
Topic: Thermodynamics
An ideal gas heat engine is operating between 227∘ C and 127∘ C. It absorbs 104 J of heat at the higher temperature. The amount of heat converted into work is
Options: (a) 2000 J
(b) 4000 J
(c) 8000 J
(d) 5600 J
Answer: 2000 J
Solution: η=1−(T 2/T 1) η=1−(127+273)/(227+273)=1−(400/500)=1/5 W=η Q 1=1/5×10 4=2000 J
Topic: Work Power Energy
Two masses of $1 \mathrm{~g}$ and $4 \mathrm{~g}$ are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is
Options: (a) $4: 1$ (b) $\sqrt{ } 2: 1$ (c) $1: 2$ (d) $1: 16$
Answer: (c) $1: 2$
Solution: $p=\sqrt{2 E m}$ Where $\mathrm{p}$ denotes momentum, $\mathrm{E}$ denotes kinetic energy and $m$ denotes mass. $$ \frac{p_{1}}{p_{2}}=\sqrt{\frac{m_{1}}{m_{2}}} \frac{p_{1}}{p_{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2} $$
Topic: Work Power Energy
If a machine is lubricated with oil
Options: (a) the mechanical advantage of the machine increases (b) the mechanical efficiency of the machine increases (c) both its mechanical advantage and efficiency increase (d) its efficiency increases, but its mechanical advantage decreases.
Answer: (b) When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.
Solution: When a machine is lubricated with oil friction decreases. Hence the mechanical efficiency of the machine increases.
Topic: Work Power Energy
The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $$ U(x)=\frac{a}{x^{12}}-\frac{b}{x^{6}} $$ between the atoms. if the dissociation energy of the molecule is $(U(x=\infty)-U$ at equilibrium), $D$ is
Options: (a) $b^{2} / 6 a$ (b) $b^{2} / 2 a$ (c) $b^{2} / 12 a$ (d) $b^{2} / 4 a$
Answer: (d) $b^{2} / 4 a$
Solution: $\mathrm{U}(\mathrm{x}=\infty)=0$, at equilibrium $\mathrm{dU} / \mathrm{dx}=0$ $U(x)=\frac{a}{x^{12}}-\frac{b}{x^{6}}--(1)$ $-12 a x^{-13}+6 b x^{-7}=0$ $-12 a / x^{13}=6 b / x^{7}$ $x^{6}=2 a / b--(2)$ From 1 and 2 we have $$ U=\frac{a-b(2 a / b)}{(2 a / b)^{2}} $$ $U=-b^{2} / 4 a$ $D=\left(0+b^{2} / 4 a\right)$ $\mathrm{D}=\mathrm{b}^{2} / 4 \mathrm{a}$
Topic: Work Power Energy
At time $t=0$ s particle starts moving along the $x$-axis. If its kinetic energy increases uniformly with time ' $t$ ', the net force acting on it must be proportional to
Options: (a) $\sqrt{ } \mathrm{t}$ (b) constant (c) $t$ (d) $1 / \sqrt{ } \mathrm{V}$
Answer: (d) $1 / \sqrt{t}$
Solution: Linear dependency with initial Kinetic Energy as zero is given as $\mathrm{KE}=\mathrm{kt}$, where $\mathrm{k}$ is the proportionality constant. Kinetic energy can be written as $\mathrm{KE}=1 / 2 \mathrm{mv}^{2}$ so we can write $1 / 2 m v^{2}=k t$ $$ v=\sqrt{\frac{2 k t}{m}} \frac{d v}{d t}=\sqrt{\frac{k}{2 m t}} $$ $F=m . d v / d t$ $$ F=m \sqrt{\frac{k}{2 m t}} $$ F $\infty 1 / \sqrt{t}$
Topic: Work Power Energy
This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. If two springs $S_{1}$ and $S_{2}$ of force constants $k_{1}$ and $k_{2}$, respectively, are stretched by the same force, it is found that more work is done on spring $\mathrm{S}_{1}$ than on spring $\mathrm{S}_{2}$. Statement-1: If stretched by the same amount, work done on $\mathrm{S}_{1}$, will be more than that on $\mathrm{S}_{2}$ Statement-2 $: k_{1}<k_{2}$
Options: (a) Statement-1 is True, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1. (b) Statement-1 is False, Statement-2 is true (c) Statement-1 is True, Statement-2 is false (d) Statement-1 is True, Statement-2 is true and Statement-2 is the correct explanation of statement-1.
Answer: (b) Statement-1 is False, Statement-2 is true
Solution: Given same force $F=k_{1} x_{1}=k_{2} x_{2}$ $k_{1} / k_{2}=x_{1} / x_{2}$ $W_{1}=1 / 2 k_{1} x_{1}^{2}$ and $W_{2}=1 / 2 k_{2} x_{2}^{2}$ As $W_{1} / W_{2}>1$ so $1 / 2 k_{1} x_{1}{ }^{2} 11 / 2 k_{2} x_{2}{ }^{2}>1$ $\mathrm{Fx}_{1} / \mathrm{Fx}_{2}>1 \Rightarrow \mathrm{k}_{2} / \mathrm{k}_{1}>1$ Therefore $k_{2}>k_{1}$ Statement- 2 is true Or if $x_{1}=x_{2}=x$ $W_{1} / W_{2}=1 / 2 k_{1} x^{2} / 1 / 2 k_{2} x^{2}$ $W_{1} / W_{2}=k_{1} / k_{2}<1$\ $\mathrm{W}_{1}<\mathrm{W}_{2}$, Statement-1 is False
Topic: Work Power Energy
A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration ac is varying with time ' $t$ ' as ac $=k^{2} r t^{2}$ where ' $k$ ' is a constant. The power delivered to the particle by the force acting on it is
Options: (a) $2 \pi \mathrm{mk}^{2} \mathrm{r}^{2} \mathrm{~s}$ (b) $m^{2} r^{2} t$ (c) $\mathrm{mk}^{4} \mathrm{r}^{2} \mathrm{t} / 3$ (d) zero
Answer: (b) $\mathrm{mk}^{2} \mathrm{r}^{2} \mathrm{t}$
Solution: Centripetal acceleration $\mathrm{a}_{\mathrm{c}}=\mathrm{k}^{2} \mathrm{rt}^{2}$ $\mathrm{V}^{2} / \mathrm{r}=\mathrm{k}^{2} \mathrm{t}^{2}$ $\mathrm{V}^{2}=\mathrm{k}^{2} \mathrm{r}^{2} \mathrm{t}^{2}$ $\mathrm{mv}^{2} / 2=\mathrm{m} \mathrm{k}^{2} \mathrm{r}^{2} \mathrm{t} 2 / 2$ Kinetic energy $=\mathrm{m}^{2} \mathrm{r}^{2} \mathrm{t}^{2} / 2$ $$ \frac{d}{d t}(K \cdot E)=m k^{2} r^{2} t $$ Power $=\mathrm{m} \mathrm{k}^{2} \mathrm{r}^{2} \mathrm{t}$
Topic: Work Power Energy
A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is
Options: (a) $\sqrt{u^{2}-2 g L}$ (b) $\sqrt{u^{2}-g L}$ (c) $\sqrt{2\left(u^{2}-g L\right)}$ (d) $\sqrt{2 g L}$
Answer: (c) $\sqrt{2\left(u^{2}-g L\right)}$
Solution: At lowest position B, Potential Energy $=0$ Kinetic Energy $=1 / 2$ mu Total Energy $=1 / 2 \mathrm{mu}$ At $\mathrm{C}$, when string is horizontal, Potential Energy $=\mathrm{mgL}$ Kinetic Energy $=1 / 2 \mathrm{mv}^{2}$ Total Energy $=1 / 2 \mathrm{mv}^{2}+\mathrm{mgL}$ Since energy is conserved, $1 / 2 m v^{2}+m g L=1 / 2 m u^{2}$ $v^{2}=u^{2}-2 g L$ Since $v$ is in vertical direction and $u$ is in horizontal direction, they are mutually perpendicular to each other. Change in velocity $=\sqrt{u^{2}+v^{2}}|\Delta \vec{v}|=$ $$ \sqrt{u^{2}+\left(u^{2}-2 g L\right)}|\Delta \vec{v}|=\sqrt{\left.2\left(u^{2}-g L\right)\right)} $$
Topic: Work Power Energy
When a rubber-band is stretched by a distance $x$, it exerts a restoring force of magnitude $F=a x+$ $b x^{2}$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is
Options: (a) $a L^{2} / 2+b L^{3} / 3$ (b) $1 / 2\left(a L^{2} / 2+b L^{3} / 3\right)$ (c) $a L^{2}+b L^{3}$ (d) $1 / 2\left(a L^{2}+b L^{3}\right)$
Answer: (a) $a L^{2} / 2+b L^{3} / 3$
Solution: Work done by a variable force $$ W=\int \vec{F} \cdot \overrightarrow{d s} $$ Wherein $\operatorname{Vvec}\{\mathrm{F}\} F$ is the variable force and $\backslash \mathrm{vec}\{\mathrm{ds}\} d s$ is small displacement $F=a x+b x^{2}$ Work done in displacing rubber through $\mathrm{dx}=\mathrm{Fdx}$ $$ W=\int_{0}^{L}\left(a x+b x^{2}\right) d x $$ $\mathrm{W}=\mathrm{aL}^{2} / 2+\mathrm{bL} 3 / 3$
Topic: Work Power Energy
A person trying to lose weight by burning fat lifts a mass of $10 \mathrm{~kg}$ up to a height of $1 \mathrm{~m} 1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \mathrm{X}$ $10^{7} \mathrm{~J}$ of energy per kg which is converted to mechanical energy with a $20 \%$ efficiency rate. Take $\mathrm{g}=$ $9.8 \mathrm{~m} / \mathrm{s}^{2}$
Options: (a) $12.89 \times 10^{-3} \mathrm{~kg}$ (b) $2.45 \times 10^{-3} \mathrm{~kg}$ (c) $6.45 \times 10^{3} \mathrm{~kg}$ (d) $9.89 \times 10^{-3} \mathrm{~kg}$
Answer: (a) $12.89 \times 10^{-3} \mathrm{Kg}$
Solution: Work done against gravity $=(\mathrm{mgh}) 1000$ in lifting 1000 times $=10 \times 9.8 \times 10^{3}$ $=9.8 \times 10^{4}$ Joule $20 \%$ of the efficiency is used to convert fat into energy $\left(20 \%\right.$ of $\left.3.8 \times 10^{7} \mathrm{~J}\right) \times \mathrm{m}=9.8 \times 10^{4}$ Where $m$ is mass $\mathrm{m}=12.89 \times 10^{-3} \mathrm{Kg}$
Topic: Work Power Energy
A body of mass $\mathrm{m}=10^{-2} \mathrm{~kg}$ is moving in a medium and experiences a frictional force $\mathrm{F}=-\mathrm{KV}^{2}$. Its initial speed is $v_{0}=10 \mathrm{~ms}^{-1}$. If, after $10 \mathrm{~s}$, its energy is, the value of $k$ will be
Options: (a) $10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$ (b) $10^{-1} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$ (c) $10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$ (d) $10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$
Answer: (a) $10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$
Solution: $1 / 2 m v_{t}^{2}=1 / 8 m v_{0}{ }^{2}$ $V_{\mathrm{f}}=\mathrm{V}_{0} / 2=5 \mathrm{~m} / \mathrm{s}$ $V_{\mathrm{f}}=\mathrm{v}_{\mathrm{o}} / 2=5 \mathrm{~m} / \mathrm{s}$ $\left(10^{-2}\right) \mathrm{dV} / \mathrm{dt}=-\mathrm{kV}^{2}$ $\int_{10}^{5} \frac{d v}{v^{2}}=-100 k \int_{0}^{10} d t$ $1 / 5-1 / 10=100 k(10)$ $\mathrm{k}=10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$
Topic: Work Power Energy
A time-dependent force $F=6 t$ acts on a particle of mass $1 \mathrm{~kg}$. If the particle starts from rest, the work done by the force during the first $1 \mathrm{sec}$. will be :
Options: (a) $9 \mathrm{~J}$ (b) $18 \mathrm{~J}$ (c) $4.5 \mathrm{~J}$ (d) $22 \mathrm{~J}$
Answer: (c) $4.5 \mathrm{~J}$
Solution: $F=6 t, m=1 \mathrm{Kg}, u=0$ Now, $F=m a=m(d v / d t)=1 x(d v / d t)$ $\mathrm{dv} / \mathrm{dt}=6 \mathrm{t}$ $\int_{0}^{v} d v=\int_{0}^{1} 6 t d t$ $v=3\left(1^{2}-0\right)=3 \mathrm{~m} / \mathrm{s}$ From work-energy theorem $W=\Delta K E=1 / 2 m\left(v^{2}-u^{2}\right)$ $W=1 / 2 \times 1\left(3^{2}-0\right)=4.5 \mathrm{~J}$
Topic: Work Power Energy
A particle is moving in a circular path of radius a under the action of an attractive potential $\mathrm{U}=$ $\mathrm{k} / 2 \mathrm{r}^{2}$.Its total energy is
Options: (a) $k / 2 a^{2}$ (b) Zero (c) $-3 / 2\left(\mathrm{k} / \mathrm{a}^{2}\right)$ (d) $-k / 4 a^{2}$
Answer: (b) Zero
Solution: $\mathrm{F}=\mathrm{mv} v^{2}=\mathrm{K} / \mathrm{r}^{2}$ $K E=1 / 2 m v^{2}=K / 2 r^{2}$ Total energy $=$ P.E + K.E $-K / 2 r^{2}+K / 2 r^{2}=$ Zero
Topic: Work Power Energy
A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of
Options: (a) $(2 / 3) \mathrm{k}$ (b) $(3 / 2) \mathrm{k}$ (c) $3 \mathrm{k}$ (d) $6 \mathrm{k}$
Answer: (b) (3/2) k
Solution: For a spring, $(K \times L)$ is a constant Therefore, $\mathrm{k} \times \mathrm{L}=\mathrm{k} \times \mathbf{x}(2 \mathrm{~L} / 3$ ) Where $2 L / 3=$ length of longer piece Or k'=(3/2)k
Topic: Work Power Energy
A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to
Options: (a) $v$ (b) $v^{2}$ (c) $v^{3}$ (d) $v^{4}$
Answer: (c) $v^{3}$
Solution: Force $=v \times d m / d t$ $$ \begin{aligned} & \text { Force }=\frac{d}{d t}(\text { volume } x \text { density })=v \frac{d}{d t}(\text { Axp }) \\ & \text { Force }=v \text { Ap } \frac{d x}{d t} \\ & \text { Power }=\text { Force } x \text { velocity } \\ & \left(\text { Apv }{ }^{2}\right) v=\mathrm{Apv}^{3} \\ & \text { Power } \propto \mathrm{V}^{3} \\ & \text { Answer: (c) } \mathbf{v}^{3} \end{aligned} $$
Topic: Centre of Mass
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
Options: (a) 30 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer: (c) 10 m/s
Solution: Just after collision
v_c = m1v1 + m2v2 / (m1 + m2)
v_c = (10 * 14 + 4 * 0) / (10 + 4)
v_c = 10 m/s
Topic: Centre of Mass
Two particles A and B, initially at rest, move towards each other under the mutual force of attraction. At the instant when the speed of A is v and the speed of B is 2v, the speed of the centre of mass of the system is
Options: (a) 3v
(b) v
(c) 1.5v
(d) zero
Answer: (d) zero
Solution: FA is the force on particle A
FA = mA * aA = mA * v / t
FB is the force on particle B
FB = mB * aB = mB * 2v / t
Since FA = FB
mA * v / t = mB * 2v / t
So mA = 2mB
For the centre of mass of the system
v = (mA * vA + mB * vB) / (mA + mB)
v = (2mB * v - mB * 2v) / (2mB + mB) = 0
The negative sign is used because the particles are travelling in the opposite directions
Topic: Centre of Mass
A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 m/s. Which of the following statements(s) are correct for the system of these two masses
Options: (a) The total momentum of the system is 3 kg m/s
(b) The momentum of 5 kg mass after the collision is 4 kg m/s
(c) The kinetic energy of the centre of mass is 0.75 J
(d) The total kinetic energy of the system is 4 J
Answer: (a) and (c)
Solution: If the velocity of 1 kg mass before the collision is u and velocity of 5 kg mass after the collision is v then
From the conservation of linear momentum
1 * u + 5 * 0 = 1 * (-2) + 5 * v
u = -2 + 5v
Since the collision is elastic, total kinetic energy will be equal
Therefore,
1/2 * 1 * u^2 = 1/2 * 1 * 2^2 + 1/2 * 5 * v^2
u^2 = 4 + 5v^2
(-2 + 5v)^2 = 4 + 5v^2
4 - 20v + 25v^2 = 0
-20v + 20v^2 = 0
-20v(1 - v) = 0
v = 1 m/s
Thus, u = 3 m/s
The velocity of the centres of mass of the combined system
V_cm = (m1v1 + m2v2) / (m1 + m2)
V_cm = (1 * (-2) + 5 * 1) / (1 + 5)
V_cm = 3 / 6
V_cm = 1 / 2
The combined Kinetic Energy of the system
Kinetic Energy = 1/2(1 + 5) * (1 / 2)^2
K.E = 6 / 8
Kinetic Energy = 0.75 J
Total momentum of the system
= (m1 + m2) * v_cm
= (1 + 5) * 1 / 2
= 3 kg-m/s
Topic: Centre of Mass
Consider a rubber ball freely falling from a height h=4.9 m onto a horizontal elastic plate. Assume that the duration of a collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be
Answer: The velocity of the ball is reversed when it strikes the surface
Solution: The velocity of the ball is reversed when it strikes the surface
Topic: Centre of Mass
Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.

Statement-2: Principle of conservation of momentum holds true for all kinds of collisions.
Options: (a) Statement-1 is true, Statement-2 is false
(b) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1
(c) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1
(d) Statement-1 is false, Statement-2 is true
Answer: (b) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1
Solution: In a completely inelastic collision then m1 v1 + m2 v2 = m1 v + m2 v
v = (m1 v1 + m2 v2) / (m1 + m2)
Kinetic Energy = p1^2 / 2 m1 + p2^2 / 2 m2
As p1 and p2 both cannot be zero simultaneously therefore total kinetic energy cannot be lost.
Topic: Centre of Mass
A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collisions are perfectly inelastic, the percentage loss in the energy during the collision is close to
Options: (a) 56%
(b) 62%
(c) 44%
(d) 50%
Answer: (a) 56%
Solution: Conservation of linear momentum in the x-direction
(pi)x = (pf)x or 2mv = (2m + m)Vx
vx = 2/3 V
Conservation of linear momentum in y-direction
(pi)y = (pf)y or 2mv = (2m + m) Vy
vy = 2/3 V
The initial kinetic energy of the two particles system is
Ei = 1/2 m (2v)^2 + 1/2 (2m) v^2
Ei = 1/2 * 4mv + 1/2 * 2mv^2
Ei = 2mv^2 + mv^2 = 3mv^2
Final energy of the combined two particles system is
Ef = 1/2 (3m) (vx^2 + vy^2)
Ef = 1/2 (3m) [4v^2 / 9 + 4v^2 / 9]
Ef = 4mv^2 / 3
Loss of energy ΔE = Ei - Ef
ΔE = mv^2 (3 - 4/3) = (5/3) mv^2
Percentage loss in the energy during the collision
ΔE / Ei * 100 = [(5/3) mv^2 / 3mv^2] * 100
= (5/9) * 100 ≈ 56%
Topic: Centre of Mass
The distance of the centre of mass of a solid uniform cone from its vertex is Z0. If the radius of its base is R and its height is h then Z0 is equal to
Options: (a) 3h / 4
(b) h^2 / 4R
(c) 5h / 8
(d) 3h^2 / 8R
Answer: (a) 3h / 4
Solution: For a solid cone centre of mass from the base is h / 4
Position of centre of mass of a solid cone from the vertex = h - h / 4 = 3h / 4