The angle of repose is defined as the maximum inclination of the plane for which the mass kept over it can stay at rest. Let Ф be the angle of repose. Then,
N = mg cosФ , F _f = mg sin Ф
Also, F _f = μN
\(\therefore \ \mu m g \cos \phi=m g \sin \phi\)
i.e., μ = tan Φ
Since the body is at rest, the friction is static friction and the coefficient is for static case.

Here, m 3.0 kg, u = 2.0 ms^-1
v = 3.5 ms^-1, t = 25 s
As F=ma
or F=m\(({v-u\over t}) [{\because a={v-u\over t}}]\)
\(\Rightarrow F={3.0(3.5-2.0)\over 25}=0.18N\)
The force is along the direction of motion.

Figure shows vectorially the situations,
 
Velocity of rain falling vertically downward v_r = 30 m/s
Velocity of woman riding a bicycle v_w=10 m/s (North to South)
To protect herself from rain, the woman should hold her umbrella in the direction of relative velocityof the rain with respect to the woman i.e.v_rw.
The relative velocity of rain with respect to the woman i.e.
v_rw=v_r-v_w
\(\left| { v }_{ rw } \right| =\sqrt { { (30) }^{ 2 }+{ (10) }^{ 2 } } \)
\(=\sqrt { 900+100 } =\sqrt { 1000 } \)
= \(10\sqrt { 10 } \) m/s
If v_rw makes an angle \(\alpha \) with the vertical, then 
\(\tan { \alpha } =\frac { { v }_{ w } }{ { v }_{ r } } =\frac { OB }{ OA } =\frac { 10 }{ 30 } \)
=\(\frac { 1 }{ 3 } \) = 0.3333
 \(\Rightarrow \alpha =18°26\prime \)
Hence, woman should hold her umbrella at an angle \(18°26\prime \) with the vertical towards South.

(a) False, because the net acceleration of a particle is towards the centre only in case of a uniform circular motion.
(b) True, because while leaving the circular path, the particle moves tangentially to the circular path.
(c) True, because the direction of acceleration vector in a uniform circular motion is directed towards the centre of circular path.It is constantly changing with time. The resultant of all these vectors will be a zero vector.

θ and (90 - θ) are two values of projection angle at which range is same, for same velocity of projection.

Linear momentum, \(\overrightarrow{p}=\overrightarrow{mv}.As \ \overrightarrow {v}=0,therefore \overrightarrow{p}=0.\)

Displacement for each girl =\(\overset\rightarrow{PQ}\)
\(\therefore\) Magnitude of the displacement for each girl 
= PQ = diameter of circular ice ground
= 2 x 200 = 400 m.

(i) T
When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,
F=T=mv² / l
Where F is the net force acting on the particle.

To better visualise the solution described here, we first sketch the trajectory as shown in figure

(i) The problem here is to find t when x = 32.0 m. We can use (x = v _xo t), if we first find v _xo. From figure, we see that v _xo = V_o cos \(\theta\)₀= (20.0 m / s) (cos 30.0°)
= 17.3 m/s
Using the relation and solve for t
x = v_xo t
\(t=\frac{x}{v_{x 0}}=\frac{32.0 \mathrm{~m}}{17.3 \mathrm{~m} / \mathrm{s}}=1.85 \mathrm{~s}\)
(ii) We want to find y when x = 32.0 m, or since we have already found the time in part (a), we can state this, find y when t = 1.85 s. Using the relation
\(y=v_{y_{0}} t-\frac{1}{2} g t^{2}\)
where v_y0 = V_o sin \(\theta\)₀ = (20.0 m /s) (sin 30.0°)
= 10.0m/s
Thus \(y=(10.0 \mathrm{~m} / \mathrm{s})(1.85 \mathrm{~s})-\frac{1}{2}\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(1.85 \mathrm{~s})^{2}\)
= 1.73 m
Since, y = 0 is 2.00 m above the ground, this means the ball is 3.73 m above the ground as it crosses the goal line too much high to be caught at that point.

The forward pulling of the tablecloth will tend to push the book backwards. The force of friction on the book will, therefore, be in the forward direction and will have a magnitude
\(F=\mu R =\mu m g\)
where m is the mass of the book. The acceleration of the book will, therefore, be
\({F\over m}={\mu m g \over m}=\mu g\)
Thus if,\(\mu\)g < a, the acceleration of the tablecloth, the cloth will slip from under the book. At the instant (say t = t), the edge of the tablecloth comes under the centre of the book, the distance of the tablecloth edge and the centre of the book from the edge of the table must be the same. Now the distance moved by the edge of the tablecloth (the initial distance from the edge of the table is zero) in a time t is
\(x'={0+{1\over2}a t^2}\)  (the tablecloth is at rest at t = 0)
The distance moved by the centre of the book in time t is
\(x'={b+{1\over2}(\mu g)t^2}\)  (acceleration of the book =\(\mu\),g)
The time instant t is, therefore, given by
\({1\over 2}a \ t^2={1\over2}\mu g t^2 +b\)
or t² (a -\(\mu\)g)=2 b
or \(t=\sqrt{2b\over (a-\mu g)}\)
The velocity of the book and the distance of its centre from the edge of the table at this instant are given by 
\(v=0+\mu g \sqrt{2b\over (a-\mu g)}\)
\(=\sqrt{2b\mu^2 g^2\over (a-\mu g)}\) 
and \(s=b+{1\over2}\mu g{2b\over(a-\mu g)}\)
\(=b\{ {1+{\mu g\over (a- \mu g)}} \}\)
\(={ab\over (a-\mu g)}\)

A) If both assertion and reason are true and the reason is the correct explanation of the assertion. 
In SHM, the acceleration is always in a direction opposite to that of the displacement i.e., proportional to (y).

B) If both assertion and reason are true but reason is not the correct explanation of the assertion.    
We know \(R=4 H \cot \theta \text { lif } R=H \text { then } \cot \theta=\left(\frac{1}{4}\right)|\operatorname{or} \tan \theta=(4)| \text { and } R=\frac{u^{2} \sin 2 \theta}{g}\left|\backslash R \propto \frac{u^{2}}{g}\right|\)