Browse and search thousands of solved questions for your preparation.
Free questions with full explanations — locked behind signup to protect quality.
A uniform force of $(3\hat{i} + \hat{j})$ N acts on a particle of mass 2 kg. The particle is displaced from the position $(2\hat{i} + \hat{k})$ m to the position $(4\hat{i} + 3\hat{j} - \hat{k})$ m. The work done by the force on the particle is:
The relation amongst the three elements of Earth's magnetic field, namely horizontal component $H$, vertical component $V$ and dip angle $\delta$ is: ($B_E$ = total magnetic field):
Given below are two statements: Assertion (A): Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. Reason (R): The magnetic monopoles do not exist. North and South poles occur in pairs, allowing vanishing net magnetic flux through the surface.
Ionized hydrogen atoms and $\alpha$-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths $r_H : r_\alpha$ will be :
A body of mass $M$ at rest explodes into three pieces, two of which of mass $M/4$ each are thrown off in perpendicular directions with velocities of $3 \text{ m/s}$ and $4 \text{ m/s}$ respectively. The third piece will be thrown off with a velocity of:
A cord is used to lower vertically a block of mass $M$ by a distance $d$ with constant downward acceleration $\frac{g}{4}$. Work done by the cord on the block is:
Match List-I with List-II. **List-I (Material)** (A) Diamagnet (B) Paramagnet (C) Soft ferromagnet (D) Hard ferromagnet **List-II (Example)** (I) Alnico (II) Copper (III) Aluminium (IV) Gadolinium Choose the correct answer from the options given below:
A block of mass $50 \text{ kg}$ slides over a horizontal distance of $1 \text{ m}$. If the coefficient of friction between their surfaces is $0.2$, then work done against friction is:
Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is
A uniform force of $(3\hat{i} + \hat{j}) \text{ N}$ acts on a particle of mass $2 \text{ kg}$. Hence the particle is displaced from position $(2\hat{i} + \hat{k}) \text{ m}$ to position $(4\hat{i} + 3\hat{j} - \hat{k}) \text{ m}$. The work done by the force on the particle is:
The cylindrical tube of a spray pump has radius $R$, one end of which has $n$ fine holes, each of radius $r$. If the speed of the liquid in the tube is $v$, the speed of the ejection of the liquid through the holes is:
Three liquids of densities $\rho_1, \rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$), having the same value of surface tension $T$, rise to the same height in three identical capillaries. The angles of contact $\theta_1, \theta_2$ and $\theta_3$ obey:
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, $1\text{ kg}$ first part moving with a velocity of $12\text{ m s}^{-1}$ and $2\text{ kg}$ second part moving with a velocity of $8\text{ m s}^{-1}$. If the third part flies off with a velocity of $4\text{ m s}^{-1}$, its mass would be:
The cylindrical tube of a spray pump has radius $R$, one end of which has $n$ fine holes, each of radius $r$. If the speed of the liquid in the tube is $v$, then the speed of ejection of the liquid through the holes will be:
The approximate depth of an ocean is $2700\text{ m}$. The compressibility of water is $45.4 \times 10^{-11}\text{ Pa}^{-1}$ and density of water is $10^3\text{ kg/m}^3$. What fractional compression of water will be obtained at the bottom of the ocean?
A wind with speed $40 \text{ m/s}$ blows parallel to the roof of a house. The area of the roof is $250 \text{ m}^2$. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: $(\rho_{air}=1.2 \text{ kg/m}^3)$
The pressure experienced by a swimmer $20 \text{ m}$ below the water surface in a lake is appropriately: (Given density of water = $10^3 \text{ kg m}^{-3}$, $g=10 \text{ m s}^{-2}$ and $1 \text{ atm} = 10^5 \text{ Pa}$)
Two particles of masses $m_1$ and $m_2$ move with initial velocities $u_1$ and $u_2$ respectively. On collision, one of the particles gets excited to a higher level, after absorbing energy $E$. If the final velocities of particles are $v_1$ and $v_2$, then we must have:
The input resistance of a silicon transistor is $100\, \Omega$. Base current is changed by $40\, \mu\text{A}$ which results in a change in collector current by $2\text{ mA}$. This transistor is used as a common-emitter amplifier with a load resistance of $4\text{ k}\Omega$. The voltage gain of the amplifier is:
The displacement of a particle executing simple harmonic motion is given by $y = A_0 + A\sin\omega t + B\cos\omega t$. Then the amplitude of its oscillation is given by :