The velocity of a particle
Question: The velocity of a particle is $\mathrm{v}=\mathrm{v}_{0}+\mathrm{gt}+\mathrm{Ft}^{2}$. Its position is $\mathrm{x}=0$ at $\mathrm{t}=0$; then its displacement after time $(\mathrm{t}=1)$ is :$v_{0}+g+F$$v_{0}+\frac{g}{2}+\frac{F}{3}$$\mathrm{v}_{0}+\frac{\mathrm{g}}{2}+\mathrm{F}$$v_{0}+2 g+3 F$Correct Option: , 2 Solution: (2) $\mathrm{v}=\mathrm{v}_{0}+\mathrm{gt}+\mathrm{Ft}^{2}$ $\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{v}_{0}+\mathrm{gt}+\mathrm{Ft}^{2}$ $\int \mathrm{ds}=\int_{0}^...
Read More →Solve this following
Question: 1 mole of rigid diatomic gas performs a work of $\mathrm{Q} / 5$ when heat $\mathrm{Q}$ is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8}$, The value of $x$ is $[\mathrm{K}=$ universal gas constant $]$ Solution: $\mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}$ $\mathrm{Q}=\Delta \mathrm{U}+\frac{\mathrm{Q}}{5}$ $\Delta U=\frac{4 Q}{5}$ $\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=\frac{4}{5} \mathrm{nC} \Delta \mathrm{T}$ $\frac{5}{4} C_{V}=C$ $...
Read More →The P-V diagram of a diatomic ideal gas system going
Question: The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process CD is (use $\gamma=1.4$ ) : $-500 \mathrm{~J}$$-400 \mathrm{~J}$$400 \mathrm{~J}$$200 \mathrm{~J}$Correct Option: 1 Solution: Adiabatic process is from $C$ to $D$ $\mathrm{WD}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}-\mathrm{P}_{1} \mathrm{~V}_{1}}{1-\gamma}$ $=\frac{200(3)-(100)(4)}{1-1.4}$ $=-500 \mathrm{~J}$...
Read More →The velocity-displacement graph describing the motion of a bicycle is shown in the figure
Question: The velocity-displacement graph describing the motion of a bicycle is shown in the figure The acceleration-displacement graph of the bicycle's motion is best described by :Correct Option: 1 Solution: For $0 \leq x \leq 200$ $\mathrm{v}=\mathrm{mx}+\mathrm{C}$ $v=\frac{1}{5} x+10$ $a=\frac{v d v}{d x}=\left(\frac{x}{5}+10\right)\left(\frac{1}{5}\right)$ $\mathrm{a}=\frac{\mathrm{x}}{25}+2 \Rightarrow$ Straight line till $\mathrm{x}=200$ for $x200$ $\mathrm{v}=\mathrm{constant}$ $\Righta...
Read More →Which one is the correct option
Question: Which one is the correct option for the two different thermodynamic processes ? (c) and (a)(c) and (d)(a) only(b) and (c)Correct Option: , 2 Solution: (2) Option (a) is wrong ; since in adiabatic process $\mathrm{V} \neq$ constant. Option (b) is wrong, since in isothermal process $\mathrm{T}=$ constant Option (c) \ (d) matches isothermes \ adiabatic formula : $\mathrm{TV}{ }^{\gamma-1}=$ constant $\ \frac{\mathrm{T}^{\gamma}}{\mathrm{p}^{\gamma-1}}=$ constant...
Read More →A sound wave of frequency
Question: A sound wave of frequency $245 \mathrm{~Hz}$ travels with the speed of $300 \mathrm{~ms}^{-1}$ along the positive $\mathrm{x}$-axis. Each point of the wave moves to and fro through a total distance of $6 \mathrm{~cm}$. What will be the mathematical expression of this travelling wave ?$\mathrm{Y}(\mathrm{x}, \mathrm{t})=0.03\left[\sin 5.1 \mathrm{x}-\left(0.2 \times 10^{3}\right) \mathrm{t}\right]$$Y(x, t)=0.06\left[\sin 5.1 \times-\left(1.5 \times 10^{3}\right) t\right]$$Y(x, t)=0.06\l...
Read More →A radioactive sample disintegrates via two independent
Question: A radioactive sample disintegrates via two independent decay processes having half lives $\mathrm{T}_{1 / 2}^{(1)}$ and $\mathrm{T}_{1 / 2}^{(2)}$ respectively. The effective half- life $\mathrm{T}_{1 / 2}$ of the nuclei is :None of the above$\mathrm{T}_{1 / 2}=\mathrm{T}_{1 / 2}^{(1)}+\mathrm{T}_{1 / 2}^{(2)}$$\mathrm{T}_{1 / 2}=\frac{\mathrm{T}_{1 / 2}^{(1)} \mathrm{T}_{1 / 2}^{(2)}}{\mathrm{T}_{1 / 2}^{(1)}+\mathrm{T}_{1 / 2}^{(2)}}$$\mathrm{T}_{1 / 2}=\frac{\mathrm{T}_{1 / 2}^{(1)}...
Read More →A geostationary satellite
Question: A geostationary satellite is orbiting around an arbitary planet 'P' at a height of $11 \mathrm{R}$ above the surface of ' $P^{\prime}$, $R$ being the radius of ' $P$ '. The time period of another satellite in hours at a height of $2 R$ from the surface of 'P' is ____________'P' has the time period of 24 hours.$6 \sqrt{2}$$\frac{6}{\sqrt{2}}$35Correct Option: , 3 Solution: (3) $\mathrm{T} \propto \mathrm{R}^{3 / 2}$ $\frac{24}{T}=\left(\frac{12 R}{3 R}\right)^{3 / 2} \Rightarrow T=3 h r...
Read More →What will be the average value of energy along
Question: What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature $T$ ? $\left(\mathrm{k}_{\mathrm{B}}\right.$ is Boltzmann constant)$\frac{1}{2} \mathrm{k}_{\mathrm{B}} \mathrm{T}$$\frac{2}{3} \mathrm{k}_{\mathrm{B}} \mathrm{T}$$\frac{3}{2} k_{B} T$$k_{\mathrm{B}} \mathrm{T}$Correct Option: 1 Solution: Energy associated with each degree of freedom per molecule $=\frac{1}{2} k_{B} T$....
Read More →Time period of a simple pendulum is T inside a lift when the lift is stationary.
Question: Time period of a simple pendulum is $\mathrm{T}$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $\mathrm{g} / 2$, the time period of pendulum will be :$\sqrt{3} \mathrm{~T}$$\frac{T}{\sqrt{3}}$$\sqrt{\frac{3}{2}} \mathrm{~T}$$\sqrt{\frac{2}{3}} \mathrm{~T}$Correct Option: , 4 Solution: When lift is stationary $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ When lift is moving upwards $\Rightarrow$ Pseudo force acts downwards $\Rightarr...
Read More →Solve this following
Question: In the reported figure of earth, the value of acceleration due to gravity is same at point A and $\mathrm{C}$ but it is smaller than that of its value at point $B$ (surface of the earth). The value of OA : AB will be $x: y$. The value of $x$ is Solution: $\mathrm{g}_{\mathrm{A}}=\frac{\mathrm{GM}(\mathrm{r})}{\mathrm{R}^{3}}$ $\mathrm{g}_{\mathrm{C}}=\frac{\mathrm{GM}}{\left(\mathrm{R}+\frac{\mathrm{R}}{2}\right)^{2}}$ $\mathrm{g}_{\mathrm{A}}=\mathrm{g}_{\mathrm{C}}$ $\frac{r}{R^{3}}=...
Read More →An object is located
Question: An object is located at $2 \mathrm{~km}$ beneath the surface of the water. If the fractional compression $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ is $1.36 \%$, the ratio of hydraulic stress to the corresponding hydraulic strain will be [Given : density of water is $1000 \mathrm{~kg} \mathrm{~m}^{-3}$ and $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$.]$1.96 \times 10^{7} \mathrm{Nm}^{-2}$$1.44 \times 10^{7} \mathrm{Nm}^{-2}$$2.26 \times 10^{9} \mathrm{Nm}^{-2}$$1.44 \times 10^{9} \mathrm{Nm}^{-2}$Co...
Read More →A constant power delivering machine has towed a box,
Question: A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time ' $t$ ' is proportional to :-$t^{2 / 3}$$\mathrm{t}^{3 / 2}$$t$$\mathrm{t}^{1 / 2}$Correct Option: , 2 Solution: $\mathrm{P}=\mathrm{C}$ $\mathrm{FV}=\mathrm{C}$ $\mathrm{M} \frac{\mathrm{dV}}{\mathrm{dt}} \mathrm{V}=\mathrm{C}$ $\frac{V^{2}}{2} \propto t$ $V \propto t^{1 / 2}$ $\frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{t}^{1 / ...
Read More →Which one of the following
Question: Which one of the following will be the output of the given circuit ? NOR GateNAND GateAND GateXOR GateCorrect Option: , 4 Solution: Conceptual...
Read More →A block of mass
Question: A block of mass $1 \mathrm{~kg}$ attached to a spring is made to oscillate with an initial amplitude of $12 \mathrm{~cm}$. After 2 minutes the amplitude decreases to $6 \mathrm{~cm}$. Determine the value of the damping constant for this motion. (take In $2=0.693$ )$0.69 \times 10^{2} \mathrm{~kg} \mathrm{~s}^{-1}$$3.3 \times 10^{2} \mathrm{~kg} \mathrm{~s} \mathrm{~s}^{-1}$$1.16 \times 10^{2} \mathrm{~kg} \mathrm{~s}^{-1}$$5.7 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$Correct Option...
Read More →Solve this following
Question: Time period of a simple pendulum is $\mathrm{T}$. The time taken to complete $5 / 8$ oscillations starting from mean position is $\frac{\alpha}{\beta} \mathrm{T}$. The value of $\alpha$ is ......... Solution: $\frac{5}{8}$ th of oscillation $=\left(\frac{1}{2}+\frac{1}{8}\right)^{\text {th }}$ of oscillation $\pi+\theta=\omega t$ $\pi+\frac{\pi}{6}=\left(\frac{2 \pi}{T}\right) \mathrm{t}$ $\frac{7 \pi}{6}=\left(\frac{2 \pi}{T}\right) t$ $t=\frac{7 \mathrm{~T}}{12}$...
Read More →The time period of a simple pendulum is given
Question: The time period of a simple pendulum is given by $T=2 \pi \sqrt{\frac{\ell}{g}}$. The measured value of the length of pendulum is $10 \mathrm{~cm}$ known to a $1 \mathrm{~mm}$ accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of ' $\mathrm{g}$ ' using this pendulum is ' $x$ '. The value of ' $x$ ' to the nearest integer is:-$2 \%$$3 \%$$5 \%$$4 \%$Correct Option: , 2 Solution: ...
Read More →If one mole of the polyatomic
Question: If one mole of the polyatomic gas is having two vibrational modes and $\beta$ is the ratio of molar specific heats for polyatomic gas $\left(\beta=\frac{C_{P}}{C_{V}}\right)$ then the value of $\beta$ is :$1.02$$1.2$$1.25$$1.35$Correct Option: , 2 Solution: (2) $f=4+3+3=10$ assuming non linear $\beta=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=1+\frac{2}{\mathrm{f}}=\frac{12}{10}=1.2$...
Read More →A conducting wire of length 'l',
Question: A conducting wire of length 'l', area of crosssection A and electric resistivity U is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :$\frac{1}{4} \frac{\mathrm{VA}}{\rho l}$$\frac{3}{4} \frac{\mathrm{VA}}{\rho l}$$\frac{1}{4} \frac{\rho l}{\mathrm{VA}}$$4 \frac{\mathrm{VA}}{...
Read More →Four identical long solenoids A,B,C and D are
Question: Four identical long solenoids $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are connected to each other as shown in the figure. If the magnetic field at the center of $\mathrm{A}$ is $3 \mathrm{~T}$, the field at the center of $\mathrm{C}$ would be : (Assume that the magnetic field is confined with in the volume of respective solenoid). $12 \mathrm{~T}$$6 \mathrm{~T}$$9 \mathrm{~T}$$1 \mathrm{~T}$Correct Option: , 4 Solution: $\phi \propto \mathrm{i}$ $\Rightarrow B \propto i$ ...
Read More →A rubber ball is
Question: A rubber ball is released from a height of $5 \mathrm{~m}$ above the floor. It bounces back repeatedly, always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )$3.0 \mathrm{~ms}^{-1}$$3.50 \mathrm{~ms}^{-1}$$2.0 \mathrm{~ms}^{-1}$$2.50 \mathrm{~ms}^{-1}$Correct Option: , 4 Solution: (4) $v_{0}=\sqrt{2 g h}$ $\mathrm{v}=\mathrm{e} \sqrt{2 \mathrm{gh}}=\sqrt{2 \mathrm{gh}}$ $\Rightarrow \mathrm{e...
Read More →27 similar drops of mercury are maintained at 10
Question: 27 similar drops of mercury are maintained at 10 $\mathrm{V}$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ............ times that of a smaller drop. Solution: $(27)\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3}$ $R=3 r$ Potential energy of smaller drop : $\mathrm{U}_{1}=\frac{3}{5} \frac{\mathrm{kq}^{2}}{\mathrm{r}}$ Potential energy of bigger drop : $U=\frac{3}{5} \frac{k Q^{2}}{R}$ $U=\frac{3}{5} \frac{k(27 q)^{...
Read More →Four equal masses, m each are placed at the corners of a square of length (l) as shown in the figure.
Question: Four equal masses, m each are placed at the corners of a square of length (l) as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be : $\mathrm{m} l^{2}$$2 \mathrm{~m} l^{2}$$3 \mathrm{~m} l^{2}$$\sqrt{3} \mathrm{ml} l^{2}$Correct Option: , 3 Solution: Moment of inertia of point mass $=$ mass $\times(\text { Perpendicular distance from axis })^{2}$ Moment of Inertia $=\mathrm{m}(0)^{2}+\mathrm{m}(l \sqrt{2})^{2}+\mathrm{m...
Read More →A thin circular ring of mass M and radius r is rotating
Question: A thin circular ring of mass $M$ and radius $r$ is rotating gabout its axis with an angular speed $\omega$. Two particles having mass $m$ each are now attached at diametrically opposite points. The angular speed of the ring will become:$\omega \frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}$$\omega \frac{\mathrm{M}+2 \mathrm{~m}}{\mathrm{M}}$$\omega \frac{M}{M+2 m}$$\omega \frac{M-2 m}{M+2 m}$Correct Option: , 3 Solution: Using conservation of angular momentum $\left(\mathrm{Mr}^{2}\right) \o...
Read More →A particle executes S.H.M. with amplitude 'a' and time period
Question: A particle executes S.H.M. with amplitude 'a' and time period $V$. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt{x} a}{2}$. The value of $x$ is Solution: $V=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}$ $\mathrm{V}_{\max }=\mathrm{A} \omega$ $\frac{\mathrm{A} \omega}{2}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}$ $\frac{\mathrm{A}^{2}}{4}=\mathrm{A}^{2}-\mathrm{x}^{2}$ $x^{2}=\frac{3 A^{2}}{4}$ $x=\frac{\sqrt{3}}{2} \mathrm{~A}$...
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