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When forces $F_1, F_2, F_3 $ are acting on a particle of mass m such that $F_2$ and $F_3$ are mutually perpendicular, then the particle remains stationary. If the force $F_1$ is now removed than the acceleration of the particle is
Here $ F_1 = F_2 +F_3 $ $ \therefore a = { F_1 \over a } $
Assertion and reason are given in following question. Each question have four options. One of them is correct select it. Assertion : Frictional forces are conservative forces. Reason : Potential energy can be associated with frictional forces
Frictional forces are non-conservative forces because they dissipate mechanical energy as heat and do not have a potential energy associated with them. Therefore, the assertion that frictional forces are conservative is false, while the reason that potential energy can be associated with frictional forces is also false. Hence, the correct option is Assertion is false. Reason is true.
Assertion and reason are given in following question. Each question have four options. One of them is correct select it. Assertion : A body of mass 1 kg is moving with an accelaration of $1ms^{-1}$ The rate of change of its momentum is 1 N. Reason : The rate of change of momentum of body = force applied on the body.
In physics, the rate of change of momentum of a body is equal to the force applied on it (Newton's Second Law). Here, for a mass of 1 kg moving with an acceleration of $1 \text{ ms}^{-2}$, the force is $ F = ma = 1 \text{ kg} \times 1 \text{ ms}^{-2} = 1 \text{ N} $. Therefore, the rate of change of its momentum is indeed 1 N. The assertion and reason are both true, and the reason correctly explains the assertion. Hence, the correct option is Assertion is true. Reason is true and reason is correct explanation for Assertion.
Assertion and reason are given in following question. Each question have four options. One of them is correct select it. Asseration : It is difficult to move bike with its breaks on. Reason : Rolling friction is converted into sliding friction, which is comparatively larger.
The assertion is true because it's indeed difficult to move a bike with its brakes on. The reason is true as well, because rolling friction, which is usually less, gets converted into sliding friction, which is higher when the brakes are applied. Therefore, the reason correctly explains the assertion.
According to newton's second low of motion, F= ma, where F is the force required to produce an accelaration a in a body of mass m. If a = 0 than F = 0. If a force acts on a body for t seconds, the effect of the force is given by impulse = $F\times t $= change in linear momentum of the body. with the help of the passage given above, choose the most appropriate alternative The force acting on a body whose linear momentum changes by $20 kgms^{-1}$ in 10 sec is
According to Newton's second law, the force acting on a body is given by the change in momentum divided by the time over which the change occurs. Here, the change in momentum is 20 kg·m/s, and the time is 10 seconds. Thus, the force is \[ F = \frac{\Delta p}{\Delta t} = \frac{20\, \text{kg·m/s}}{10\, \text{s}} = 2\, \text{N} \].
What is the maximum value of the force f such that the block shown in the arrangement does not move
Coefficient of friction =1/(2√3)
Match the column Column - I (a) Body lying on a horizontal surface (b) Static friction (c) Limiting friction (d) Dynamic friction Column - II (p) is a self adjusting force (q) is a maximum value of static friction (r) is less than limiting friction (s) force of friction = 0
Matching the columns correctly: (a) Body lying on a horizontal surface - (s) force of friction = 0 (b) Static friction - (p) is a self-adjusting force (c) Limiting friction - (q) is a maximum value of static friction (d) Dynamic friction - (r) is less than limiting friction
An element $ \vec dl = dx \uparrow $ (where dx = 1 cm) is placed at the origin and carries a large current I = 10 Amp. What is the mag. field on the Y-axis at a distance of 0.5 meter ?
$ d l = dx = 10^{-2} m $ I = 10 Amp ; r =0.5 m $ d \vec B = { \mu_0 \over 4 \pi } { I \vec dl \times \vec r \over r^3 } $ $ = 4 \times 10^{-8} \hat k\; tesla $
Two straight long conductors AOB and COD are perpendicular to each other and carry currents $I_1$ and $I_2$. The magnitude of the mag. field at a point "P" at a distance "a" from the point "O" in a direction perpendicular to the plane ABCD is
Point "P" is lying symmetricallyw.r.t.the two long wires $ B_1 = { \nu_0 \over 2 \pi } { I_1 \over a } ;$ $ B_2 = { \mu_0 \over 2 \pi} { I_2 \over a } $ $ B = \sqrt { B_1^2 + B_1^2 } $ $ = { \mu_0 \over 2 \pi a } ( I_1^2 + I_2^2 )^{1/2} $
A length L of wire carries a steady current 1. It is bent first to form a coil of 1 turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is.................
$ B = N \left( {\mu_0 I \over 2 \pi } \right) \Rightarrow B \alpha {N \over r } \Rightarrow {B_1 \over B_2 } = { N_1 \over r_1} \times { r_2 \over N_2 }$ $ B_2 = 4 B_1 $ Technique $ B_2 = n^2 B1 = (2)^2 B1 = 4B1 $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.