NLM

Let $a_{2}$ be acceleration of m w.r.t. M
Then, $ma_{1}\cos\theta - mg\sin\theta = ma_{2}$
$\frac {a _ {1}}{\sqrt {2}} - \frac {g}{\sqrt {2}} = a _ {2}$
$\Rightarrow a _ {1} - g = a _ {2} \sqrt {2}\tag{... (1}$
Also, $N = ma_{1}\sin \theta +mg\cos \theta$
$\ldots(2)$
and $F - N\sin \theta = Ma_{1}$
Put (2) in (3), we get
$F - m a _ {1} \sin^ {2} \theta - m g \sin \theta \cos \theta = M a _ {1}\tag{... (3}$
$\Rightarrow F - m g \sin \theta \cos \theta = (M + m \sin^ {2} \theta) a _ {1}$
$\Rightarrow 2 1 0 - 2 0 \left(\frac {1}{2}\right) = \left[ 9 + 2 \left(\frac {1}{2}\right) \right] a _ {1}$
$\Rightarrow a_{1} = 20 \mathrm{~ms}^{-2}$
Hence, the correct answer is (C).

Since, $AB = l = \frac{1}{2} a_{2} t^{2}$ , where $a_{2} = \frac{a_{1} - g}{\sqrt{2}} = \frac{20 - 10}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{ms}^{-2}$ $\Rightarrow \quad \frac{10}{\cos(45^{\circ})} = \frac{1}{2} \left( \frac{10}{\sqrt{2}} \right) t^{2}$ $\Rightarrow \quad 10\sqrt{2} = \frac{10}{2\sqrt{2}} t^{2}$ $\Rightarrow \quad t = 2 \, \text{s}$ Since $v = u + at$ $\Rightarrow \quad v = a_{2} t = \left( \frac{10}{\sqrt{2}} \right)^{2} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{ms}^{-1}$
Hence, the correct answer is (C).

$l'=\frac{1}{2}a_{1}t^{2}$ $\Rightarrow \quad l'=\frac{1}{2}(20)(4)=40\ m$ Hence, the correct answer is (D).

$F\sin \theta +f_{2} = mg$ $\Rightarrow F\sin \theta +\mu F\cos \theta = mg$ $\Rightarrow F = \frac{mg}{\sin\theta + \mu\cos\theta}$ Hence, the correct answer is (B).

$F\sin \theta = mg + f_{2}$ $\Rightarrow F\sin \theta -\mu F\cos \theta = mg$ $\Rightarrow F = \frac{mg}{\sin\theta - \mu\sin\theta}$ Hence, the correct answer is (C).

$F\sin \theta = mg$ Hence, the correct answer is (B).

Since there is no force on upper block of mass 10 kg and no tendency of slipping over 20 kg block therefore friction on it will be zero.
Hence, the correct answer is (A).

Maximum acceleration or retardation upto which both blocks will move together is
$\begin{array}{r l} & a _ {\max} = \mu_ {1} g = 5 \mathrm{ms} ^ {- 2} \\ \Rightarrow & \mu_ {2} (m _ {1} + m _ {2}) g = (m _ {1} + m _ {2}) a \\ \Rightarrow & \mu_ {2} = 0. 5 \end{array}$
Also kinetic friction on 20 kg is
$f _ {k} = \mu_ {2} (m _ {1} + m _ {2}) g = 1 5 0 \mathrm{N}$
Hence, the correct answer is (A).

Since, retardation is $5 \, ms^{-2}$ , so applying
$v ^ {2} = u ^ {2} - 2 a s, \mathrm{weget}$
$0 ^ {2} = 4 0 0 - 2 \times 5 \times s$
$\Rightarrow s = 1 0 0 \mathrm{m}$
Hence, the correct answer is (B).

$\omega = at^{2}$
Linear speed $v = \omega r = 1 \times 2 \times t^{2} = (2 \, \text{ms}^{-3}) (t^{2})$
Since, $a_{T} = \frac{dv}{dt} = \frac{d(2t^{2})}{dt} = 4t$ and
$a _ {C} = \frac {v ^ {2}}{r} = \omega^ {2} r$
Net acceleration is
$a = a _ {\mathrm{net}} = \sqrt {a _ {T} ^ {2} + a _ {C} ^ {2}}$
$\mathrm{Force} = ma = m\sqrt{a_T^2 + a_C^2}$
Hence, the correct answer is (B).

A

Since rate of change of speed is tangential acceleration, so we have
$\begin{array}{r l} & a _ {T} = \frac {a}{\sqrt {2}} \\ \Rightarrow & a _ {T} ^ {2} + a _ {C} ^ {2} = a ^ {2} \\ \Rightarrow & \frac {a ^ {2}}{2} + a _ {C} ^ {2} = a ^ {2} \\ \Rightarrow & a _ {C} = \frac {a}{\sqrt {2}} \\ \Rightarrow & a _ {C} = a _ {T} = \frac {a}{\sqrt {2}} \\ \Rightarrow & 4 t = 4 t ^ {3} \\ \Rightarrow & t = 1 \mathrm{s} \end{array}$
Hence, the correct answer is (B).

The correct answer is (B).

The correct answer is (B).

The correct answer is (C).
Combined solution to 67, 68, 69

For equilibrium of
8 kg block, $T_{1} \leq 40$ N
6 kg block, $T_{2} \leq 24$ N
Also, $T_{1} = \frac{T}{2}$ and $T_{2} = \frac{T}{4}$
So, for equilibrium of
8 kg block, we have $T \leq 80$ N and
6 kg block, we have $T \leq 96$ N
However, system will be equilibrium if all the three blocks are in equilibrium. Hence
$\begin{array}{r l} & T \leq 8 0 \mathrm{N} \\ \Rightarrow & m \leq 8 \mathrm{kg} \end{array}$
The 6 kg block will start moving if
$\begin{array}{r l} & T _ {2} > 9 6 \mathrm{N} \\ \Rightarrow & m > 9. 6 \mathrm{kg} \end{array}$
When 9 kg block is hanged, 6 kg block remains at rest, but 8 kg block moves. Let acceleration of 9 kg block be a, then acceleration of 8 kg block will be 2a. For 9 kg block, we have
9 0 - T = 9 a
For 8 kg block, we have
... (1)
$\begin{array}{l l} & \frac {T}{2} - f = 8 (2 a) \\ \Rightarrow & \frac {T}{2} - 4 0 = 8 (2 a) \end{array}\tag{... (2}$
From (1) and (2), we get
$\begin{array}{r l} & {\frac {9 0 - 9 a}{2} = 1 6 a + 4 0} \\ {\Rightarrow} & {a = \frac {1 0}{4 1} \mathrm{ms} ^ {- 2}} \end{array}$

$N\sin \theta = kx$
$N = mg\cos \theta$
$\Rightarrow x = \frac {N \sin \theta}{k} = \frac {m g \sin \theta \cos \theta}{k} = \frac {m g \sin (2 \theta)}{2 k}$
Hence, the correct answer is (C).

$N \cos \theta + Mg = N_{1}$
$\Rightarrow N _ {1} = (m g \cos \theta) \cos \theta + M g$
$\Rightarrow N _ {1} = m g \cos^ {2} \theta + M g$
$\Rightarrow N _ {1} = (m \cos^ {2} \theta + M) g$
Hence, the correct answer is (D).

Net unbalanced force on the block is $mg \sin \theta$ . Hence, the correct answer is (A).

Friction force will be static in nature. So $f = 30 \mathrm{~N}$ Hence, the correct answer is (C).

Slipping will start if acceleration a of system is
$\begin{array}{r l} & a > \frac {0 . 3 \times 2 5 \times 1 0}{5} \\ \Rightarrow & \left(\frac {M}{M + 3 0}\right) g > \frac {0 . 3 \times 2 5 \times 1 0}{5} \end{array}$
Hence, the correct answer is (D).

$a = \left(\frac{M}{M + 30}\right)g = 6\mathrm{ms}^{-2}$
Hence, the correct answer is (A).

$v = kt_{1}$
Tangential acceleration is
$a _ {T} = \frac {d v}{d t} = k$
Centripetal acceleration is
$a _ {C} = \frac {v ^ {2}}{r}$
$\Rightarrow \frac {v ^ {2}}{r} = k$
$\Rightarrow \frac {k ^ {2} t _ {1} ^ {2}}{r} = k$
$\Rightarrow t _ {1} = \sqrt {\frac {r}{k}}$
Hence, the correct answer is (C).

Net force $F = \sqrt{mk^{2} + mk^{2}} = \sqrt{2}mk$ Hence, the correct answer is (B).

$\tan\theta=\frac{a_{T}}{a_{C}}=\frac{K}{\left(\frac{v^{2}}{r}\right)}=\frac{K}{\left(\frac{a_{T}^{2}t^{2}}{r}\right)}$
$\Rightarrow \tan \theta = \frac {K}{\left(\frac {4}{3}\right) \frac {K ^ {2} t _ {1} ^ {2}}{r}} = \frac {3}{4}$
$\Rightarrow \tan \theta = \frac {3}{4}$
$\Rightarrow \theta = 3 7 ^ {\circ}$
Hence, the correct answer is (B).

Free body diagram of block

$\Rightarrow a _ {\mathrm{block}} = \frac {1 6 0}{1 6 0} = 1 \mathrm{ms} ^ {- 2}$
{leftwards}
Free body diagram (platform + mass) system
Assuming no friction between platform and surface, we get acceleration of pulley to be equal to acceleration of platform. So

$a _ {\mathrm{pulley}} = \frac {1 6 0}{8 0} = 2 \mathrm{ms} ^ {- 2}$
{rightwards}
Relative acceleration of pulley and block is
$(2 + 1) = 3 \mathrm{ms} ^ {- 2}$
So, time taken by block to reach pulley is
$l = \frac {1}{2} a _ {\mathrm{rel}} t ^ {2}$
$\Rightarrow t = \sqrt {\frac {2 \times 6}{3}} = 2 \mathrm{s}$
Hence, the correct answer is (B).