32.
A string of negligible mass going over a damped pulley of mass $$m$$ supports a block of mass $$M$$ as shown in the figure. The force on the pulley by the clamp is given by
A
$$\sqrt 2 \,{\text{Mg}}$$
B
$$\sqrt 2 \,{\text{mg}}$$
C
$$\sqrt {{{\left( {M + m} \right)}^2} + {m^2}} g$$
D
$$\sqrt {{{\left( {M + m} \right)}^2} + {M^2}} g$$
At equilibrium $$T = Mg$$
F.B.D. of pulley
$${F_1} = \left( {m + M} \right)g$$
The resultant force on pulley is
$$F = \sqrt {F_1^2 + {T^2}} = \left[ {\sqrt {{{\left( {m + M} \right)}^2} + {M^2}} } \right]g$$
33.
Two blocks of mass $${M_1} = 20\,kg$$ and $${M_2} = 12\,kg$$ are connected by a metal rod of mass $$8\,kg.$$ The system is pulled vertically up by applying a force of $$480\,N$$ as shown. The tension at the mid-point of the rod is :
Acceleration produced in upward direction
$$\eqalign{
& a = \frac{F}{{{M_1} + {M_2} + {\text{Mass}}\,{\text{of}}\,{\text{metal}}\,{\text{rod}}}} \cr
& = \frac{{480}}{{20 + 12 + 8}} = 12\,m{s^{ - 2}} \cr} $$
Tension at the mid point
$$\eqalign{
& T = \left( {{M_2} + \frac{{{\text{Mass}}\,{\text{of}}\,{\text{rod}}}}{2}} \right)a \cr
& = \left( {12 + 4} \right) \times 12 = 192\,N \cr} $$
34.
Two forces are such that the sum of their magnitudes is $$18N$$ and their resultant is $$12N$$ which is perpendicular to the smaller force. Then the magnitudes of the forces are
Let the two forces be $${F_1}$$ and $${F_2}$$ and let $${F_2} < {F_1}.\,R$$ is the resultant force.
Given $${F_1} + {F_2} = 18\,......\left( {\text{i}} \right)$$
From the figure $$F_2^2 + {R_2} = F_1^2$$
$$\eqalign{
& F_1^2 - F_2^2 = {R^2} \cr
& \therefore F_1^2 - F_2^2 = 144\,......\left( {{\text{ii}}} \right) \cr} $$
Only option (B) follows equation (i) and (ii).
35.
Two particles of mass $$m$$ each are tied at the ends of a light string of length $$2a.$$ The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $$'a'$$ from the centre $$P$$ (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $$F.$$ As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $$2x,$$ is
A
$$\frac{F}{{2m}}\frac{a}{{\sqrt {{a^2} - {x^2}} }}$$
B
$$\frac{F}{{2m}}\frac{x}{{\sqrt {{a^2} - {x^2}} }}$$
C
$$\frac{F}{{2m}}\frac{x}{a}$$
D
$$\frac{F}{{2m}}\frac{{\sqrt {{a^2} - {x^2}} }}{x}$$
The acceleration of mass $$m$$ is due to the force $$T\cos \theta $$
$$\eqalign{
& \therefore T\cos \theta = ma \Rightarrow a = \frac{{T\cos \theta }}{m}\,......\left( {\text{i}} \right) \cr
& {\text{also, }}F = 2T\sin \theta \Rightarrow T = \frac{F}{{2\sin \theta }}......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii)
$$\eqalign{
& a = \left( {\frac{F}{{2\sin \theta }}} \right)\frac{{\cos \theta }}{m} \cr
& = \frac{F}{{2m\tan \theta }} = \frac{F}{{2m}}\frac{x}{{\sqrt {{a^2} - {x^2}} }}\,\,\left[ {\because \tan \theta \frac{{\sqrt {{a^2} - {x^2}} }}{x}} \right] \cr} $$
36.
A marble block of mass $$2kg$$ lying on ice when given a velocity of $$6m/s$$ is stopped by friction in $$10s.$$ Then the coefficient of friction is
$$\eqalign{
& u = 6{\text{m/s}},v = 0,t = 10s \cr
& a = - \frac{f}{m} = \frac{{ - \mu mg}}{m} = - \mu g = - 10\mu \cr
& v = u + at \cr
& 0 = 6 - 10\mu \times 10 \cr
& \therefore \mu = 0.06 \cr} $$
37.
A body, under the action of a force $$F = 6\hat i - 8\hat j + 10\hat k,$$ acquires an acceleration of $$1\,m{s^{ - 2}}.$$ The mass of this body must be
$$\eqalign{
& {\text{Here,}}\,F = 6\hat i - 8\hat j + 10\hat k \cr
& \left| F \right| = \sqrt {36 + 64 + 100} \cr
& = 10\sqrt 2 \,N \cr
& a = 1\,m{s^{ - 2}} \cr
& \therefore m = \frac{{10\sqrt 2 }}{1}\,\,\left[ {\because F = ma} \right] \cr
& = 10\sqrt 2 \,kg \cr} $$
38.
One end of a massless rope, which passes over a massless and frictionless pulley $$P$$ is tied to a hook $$C$$ while the other end is free. Maximum tension that the rope can bear is $$360\,N.$$ With what value of maximum safe acceleration $$\left( {{\text{in}}\,m{s^{ - 2}}} \right)$$ can a man of $$60\,kg$$ climb on the rope?
Supposing block is fixed and wedge moves with velocity $$3m/s$$ relative to block, and so
$$\eqalign{
& 3\sin {30^ \circ } = {v_y}\cos {30^ \circ }, \cr
& \therefore {v_y} = \sqrt 3 \,m/s \cr
& {\text{and}}\,{u_x} = \frac{1}{{\sin {{30}^ \circ }}} = 2\,m/s \cr
& \therefore u = \sqrt {u_x^2 + u_y^2} = \sqrt {4 + 3} = \sqrt 7 \,m/s. \cr} $$
40.
Two blocks $$A$$ and $$B$$ of masses $$3\,m$$ and $$m$$ respectively are connected by a massless and inextensible string. The whole inextensible string. The whole massless spring as shown in figure.
The magnitudes of acceleration of $$A$$ and $$B$$ immediately after the string is cut, are respectively :