When vector $\vec { A } = 2 \hat { i } + 3 \hat { j } + 2 \widehat { k }$ is subtracted from vector $\vec { B }$, it gives a vector equal to $2 \hat { j }$. Then the magnitude of vector $\vec { B }$ will be: (1) $\sqrt { 5 }$ (2) 3 (3) $\sqrt { 6 }$ (4) $\sqrt { 33 }$
A body of mass 500 g moves along $x$-axis such that it's velocity varies with displacement $x$ according to the relation $v = 10 \sqrt { x } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the force acting on the body is: (1) 125 N (2) 25 N (3) 166 N (4) 5 N
A block of mass 5 kg starting from rest pulled up on a smooth incline plane making an angle of $30 ^ { \circ }$ with horizontal with an effective acceleration of $1 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The power delivered by the pulling force at $t = 10 \mathrm {~s}$ from the start is $\_\_\_\_$ W. [Use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$] (Calculate the nearest integer value)
A circular plate is rotating in horizontal plane, about an axis passing through its centre and perpendicular to the plate, with an angular velocity $\omega$. A person sits at the centre having two dumbbells in his hands. When he stretched out his hands, the moment of inertia of the system becomes triple. If $E$ be the initial Kinetic energy of the system, then final Kinetic energy will be $\frac { E } { x }$. The value of $x$ is
A space ship of mass $2 \times 10 ^ { 4 } \mathrm {~kg}$ is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and radius of earth $= 6400 \mathrm {~km}$): (1) $11.2 ( \sqrt { 2 } - 1 ) \mathrm { km } \mathrm { s } ^ { - 1 }$ (2) $8 ( \sqrt { 2 } - 1 ) \mathrm { km } \mathrm { s } ^ { - 1 }$ (3) $7.9 ( \sqrt { 2 } - 1 ) \mathrm { km } \mathrm { s } ^ { - 1 }$ (4) $7.4 ( \sqrt { 2 } - 1 ) \mathrm { km } \mathrm { s } ^ { - 1 }$
The surface tension of soap solution is $3.5 \times 10 ^ { - 2 } \mathrm {~N} \mathrm {~m} ^ { - 1 }$. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is $\_\_\_\_$ $\times 10 ^ { - 4 } \mathrm {~J}$. $\left($ take $\left. \pi = \frac { 22 } { 7 } \right)$