A current of 1.5 A is flowing through a triangle,
Question: A current of $1.5 \mathrm{~A}$ is flowing through a triangle, of side $9 \mathrm{~cm}$ each. The magnetic field at the centroid of the triangle is: (Assume that the current is flowing in the clockwise direction.)$3 \times 10^{-7} \mathrm{~T}$, outside the plane of triangle$2 \sqrt{3} \times 10^{-7} \mathrm{~T}$, outside the plane of triangle$2 \sqrt{3} \times 10^{-5} \mathrm{~T}$, inside the plane of triangle$3 \times 10^{-5} \mathrm{~T}$, inside the plane of triangleCorrect Option: , ...
Read More →Four identical hollow cylindrical columns of mild steel support
Question: Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 \times 10^{3} \mathrm{~kg}$, The inner and outer radii of each column are $50 \mathrm{~cm}$ and $100 \mathrm{~cm}$ respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use $\mathrm{Y}=2.0 \times 10^{11} \mathrm{~Pa}, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ ]$3.60 \times 10^{-8}$$2.60 \times 10^{-7}$$1.87 \times 10^{-3}$$7.07 \times 10^{-4}$Co...
Read More →An engine is attached to a wagon through a shock absorber of length
Question: An engine is attached to a wagon through a shock absorber of length $1.5 \mathrm{~m}$. The system with a total mass of $40,000 \mathrm{~kg}$ is moving with a speed of $72 \mathrm{kmh}^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by $1.0 \mathrm{~m}$. If $90 \%$ of energy of the wagon is lost due to friction, the spring constant is $\times 10^{5} \mathrm{~N} / \mathrm{m}$. Solu...
Read More →A carrier wave with amplitude of
Question: A carrier wave with amplitude of $250 \mathrm{~V}$ is amplitude modulated by a sinusoidal base band signal of amplitude $150 \mathrm{~V}$. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is $50: \mathrm{x}$, then value of $x$ is ........... Solution: $A_{\max }=A_{C}+A_{m}=250+150=400$ $A_{\min }=A_{C}-A_{m}=250-150=100$ $\frac{A_{\min }}{A_{\max }}=\frac{100}{400}=\frac{1}{4}=\frac{50}{200}$ $x=200$...
Read More →Solve this following
Question: A $2 \mathrm{~kg}$ steel rod of length $0.6 \mathrm{~m}$ is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is ...........ms ${ }^{-1}$. (Take $g=10 \mathrm{~ms}^{-2}$ ) Solution: by energy conservation $\mathrm{mg} \ell=\frac{1}{2} \mathrm{I} \omega^{2}=\f...
Read More →There are two infinitely long straight current carrying conductors
Question: There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is ____. $\frac{\mu_{0} I}{4 \pi x y}\left[\sqrt{x^{2}+y^{2}}+(x+y)\right]$$\frac{\mu_{0} I}{4 \pi x y}\left[\sqrt{x^{2}+y^{2}}-(x+y)\right]$$\frac{\mu_{0} \mathrm{Ixy}}{4 \pi}\left[\sqrt{\mathrm{x}^{2}+\mathrm{y}^{...
Read More →Solve this following
Question: A steel rod with $\mathrm{y}=2.0 \times 10^{11} \mathrm{Nm}^{-2}$ and $\alpha=10^{-5}{ }^{\circ} \mathrm{C}^{-1}$ of length $4 \mathrm{~m}$ and area of cross-section $10 \mathrm{~cm}^{2}$ is heated from $0^{\circ} \mathrm{C}$ to $400^{\circ} \mathrm{C}$ without being allowed to extend. The tension produced in the rod is $x \times 10^{5} \mathrm{~N}$ where the value of $\mathrm{x}$ is ............. Solution: Thermal force $F=A y \propto \Delta T$ $\mathrm{F}=\left(10 \times 10^{-4}\righ...
Read More →A uniform heating wire of resistance
Question: A uniform heating wire of resistance $36 \Omega$ is connected across a potential difference of $240 \mathrm{~V}$. The wire is then cut into half and potential difference of $240 \mathrm{~V}$ is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be $1: \mathrm{x}$, where $\mathrm{x}$ is.......... Solution: First case $\mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{(240)^{2}}{36}$ Second case R...
Read More →Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by
Question: Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by $E=20 \cos \left(2 \times 10^{10} \mathrm{t}-200 \mathrm{x}\right) \mathrm{V} / \mathrm{m}$. The dielectric constant of the medium is equal to : $\left(\right.$ Take $\mu_{\mathrm{r}}=1$ )92$\frac{1}{3}$3Correct Option: 1 Solution: Speed of wave $=\frac{2 \times 10^{10}}{200}=10^{8} \mathrm{~m} / \mathrm{s}$ Refractive index $=\frac{3 \times 10^{8}}{10^{8}}=3$ Now refractive index $=\sqrt...
Read More →The average translational kinetic energy of
Question: The average translational kinetic energy of $\mathrm{N}_{2}$ gas molecules at $.{ }^{\circ} \mathrm{C}$ becomes equal to the K.E. of an electron accelerated from rest through a potential difference of $0.1$ volt. $\left(\right.$ Given $\left.\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right)$ (Fill the nearest integer). Solution: Given Translation K.E. of $\mathrm{N}_{2}=\mathrm{K}$.E. of electron $\frac{3}{2} \mathrm{kT}=\mathrm{eV}$ $\frac{3}{2} \times 1.38 ...
Read More →Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is :
Question: Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is : $\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}(2 \sqrt{2}+1)}}$$\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(2 \sqrt{2}+1)}$$\frac{1}{2} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(2 \sqrt{2}-1)}$$\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$Correct Option: , 2 Solution: $\mathrm{F}_{\mathrm{net}}=\frac{\mathrm{MV}^{2}}{\ma...
Read More →When a body slides down from rest along a smooth inclined plane making an angle
Question: When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal, it takes time $\mathrm{T}$. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time $\alpha \mathrm{T}$, where $\alpha$ is a constant greater than 1. The co-efficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ wher...
Read More →A capacitor is connected to a 20 V battery through a resistance
Question: A capacitor is connected to a $20 \mathrm{~V}$ battery through a resistance of $10 \Omega$. It is found that the potential difference across the capacitor rises to $2 \mathrm{~V}$ in $1 \mu \mathrm{s}$. The capacitance of the capacitor is___________ $\mu \mathrm{F} .$ Given : $\ln \left(\frac{10}{9}\right)=0.105$$9.52$$0.95$$0.105$$1.85$Correct Option: , 2 Solution: $\mathrm{V}=\mathrm{V}_{0}\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\right)$ $2=20\left(1-\mathrm{e}^{-\mathrm{t} / \...
Read More →Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction.
Question: Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is $2 \times 10^{3} \mathrm{~km}$. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{\pi}{x} \operatorname{rad~h}^{-1}$ where $x$ is .......... Solution: $\mathrm{T}_{1}=1$ hour $\Rightarrow \o...
Read More →The width of one of the two slits in a Young's double slit experiment is three times the other slit.
Question: The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is $x: 4$ where $x$ is Solution: Given amplitude $\propto$ slit width Also intensity $\propto(\text { Amplitude })^{2} \propto(\text { Slit width })^{2}$ $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{3}{1}\right)^{2}=9 \Rightarrow...
Read More →A mass of 5 kg is connected to a spring.
Question: A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed? $10 \mathrm{~m} / \mathrm{s}^{2}$$5 \mathrm{~m} / \mathrm{s}^{2}$$4 \mathrm{~m} / \mathrm{s}^{2}$$9.8 \mathrm{~m} / \mathrm{s}^{2}$Correct Option: , 3 Sol...
Read More →A car is moving on a plane inclined at
Question: A car is moving on a plane inclined at $30^{\circ}$ to the horizontal with an acceleration of $10 \mathrm{~ms}^{-2}$ parallel to the plane upward. A bob is suspended by a string from the roof of the car.The angle in degrees which the string makes with the vertical is (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ) Solution: $\tan (30+\theta)=\frac{m g \sin 30^{\circ}+m a}{m g \cos 30^{\circ}}$ $\tan (30+\theta)=\frac{5+10}{5 \sqrt{3}}=\frac{1+2}{\sqrt{3}}$ $\frac{\tan \theta+\frac{1}{\sqrt{3...
Read More →A capacitor of
Question: A capacitor of $50 \mu \mathrm{F}$ is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is__________ $\mu \mathrm{C} .$ Solution: Pot. Diff. across each resistor $=2 \mathrm{~V}$ $\mathrm{q}=\mathrm{CV}$ $=50 \times 10^{-6} \times 2=100 \times 10^{-6}=100 \mu \mathrm{C}$...
Read More →Solve this following
Question: The temperature of $3.00 \mathrm{~mol}$ of an ideal diatomic gas is increased by $40.0^{\circ} \mathrm{C}$ without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $\frac{x}{10}$. Then the value of $x$ (round off to the nearest integer) is (Given $\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ) Solution: Pressure is not changing $\Righta...
Read More →The electric field in an electromagnetic wave
Question: The electric field in an electromagnetic wave is given by $\mathrm{E}=\left(50 \mathrm{NC}^{-1}\right) \sin \omega(\mathrm{t}-\mathrm{x} / \mathrm{c})$ The energy contained in a cylinder of volume $V$ is $5.5 \times 10^{-12} \mathrm{~J}$. The value of $\mathrm{V}$ is $\mathrm{cm}^{3}$. Solution: $\mathrm{E}=50 \sin \left(\omega \mathrm{t}-\frac{\omega}{\mathrm{c}} \cdot \mathrm{x}\right)$ Energy density $=\frac{1}{2} \in_{0} \mathrm{E}_{0}^{2}$ Energy density $=\frac{1}{2} \in_{0} \mat...
Read More →An object of mass 'm' is being moved with a constant velocity under the action of an applied force of 2N along a frictionless surface with following surface profile.
Question: An object of mass 'm' is being moved with a constant velocity under the action of an applied force of 2N along a frictionless surface with following surface profile. The correct applied force vs distance graph will be:Correct Option: , 2 Solution: During upward motion F = 2N = (+ve) constant During downward motion $\Rightarrow \mathrm{F}=2 \mathrm{~N}=(-\mathrm{ve})$ constant $\Rightarrow$ Best possible answer is option (2)...
Read More →A block moving horizontally on a smooth surface
Question: A block moving horizontally on a smooth surface with a speed of $40 \mathrm{~ms}^{-1}$ splits into two equal parts. If one of the parts moves at $60 \mathrm{~ms}^{-1}$ in the same direction, then the fractional change in the kinetic energy will be $x: 4$ where $x=$________. Solution: $P_{i}=P_{f}$' $\mathrm{m} \times 40=\frac{\mathrm{m}}{2} \times \mathrm{v}+\frac{\mathrm{m}}{2} \times 60$ $40=\frac{V}{2}+30$ $\Rightarrow v=20$ $(\mathrm{K} . \mathrm{E} .)_{1}=\frac{1}{2} \mathrm{~m} \...
Read More →For the given circuit the current i through the battery when the key in closed
Question: For the given circuit the current i through the battery when the key in closed and the steady state has been reached is_________. 6 A25 A10 A0 ACorrect Option: , 3 Solution: In steady state, inductor behaves as a conducting wire. So, equivalent circuit becomes $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$ $\Rightarrow R_{\text {eq }}=1 \Omega$ $\Rightarrow$ Circuit becomes $\Rightarrow \mathrm{i}=\frac{30}{3}=10 \mathrm{~A}$...
Read More →In the given figure, each diode has a forward bias resistance
Question: In the given figure, each diode has a forward bias resistance of $30 \Omega$ and infinite resistance in reverse bias. The current $I_{1}$ will be : 3.75 A2.35 A2 A2.73 ACorrect Option: , 3 Solution: As per diagram, Diode D1 D2 are in forward bias i.e. R = 30: whereas diode D3 is in reverse bias i.e. R = infinite $\Rightarrow$ Equivalent circuit will be Applying KVL starting from point A $-\left(\frac{I_{1}}{2}\right) \times 30-\left(\frac{I_{1}}{2}\right) \times 130-I_{1} \times 20+200...
Read More →The voltage drop across
Question: The voltage drop across $15 \Omega$ resistance in given figure will be $\mathrm{V}$________. Solution: $\Rightarrow$ effective circuit diagram will be Point drop across $6 \Omega=1 \times 6=6=\mathrm{V}_{\mathrm{AB}}$ $\Rightarrow$ Hence point drop across $15 \Omega=6$ volt $=\mathrm{V}_{\mathrm{AB}}$...
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