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CBSE Questions for Class 8 General Knowledge Basic Science Quiz 8 - MCQExams.com
CBSE
Class 8 General Knowledge
Basic Science
Quiz 8
The following figure shows an accelerations conveyor belt in client at an angle 37 degree above horizontal. the coefficient of friction between the belt of block is $$'I'$$ . the least time in which block can reach the top starting from rest at the bottom is
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2s
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4s
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1s
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0.5s
Explanation
We know that
$$ S = 1/2 at^2$$
$$ \dfrac{4×2}{a} = t^2 $$
$$ \dfrac{8}{a} = 4$$
$$t= 2sec$$
A man moves in an open field such that after moving 10 m on a straight line, he makes a sharp turn of $$60^{0}$$ to his left. The total displacement just at the start of $$8^{th}$$ turn is equal to
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12 m
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15 m
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17.32 m
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14.14 m
A small sphere of mass $$1\ g $$ rests at one cubical box. The box is moving as such is position is $$r = 4t^{2} \hat{i} - \dfrac{2}{3} t^{3} \hat{j} $$. The force exerted on the ball at $$t = 1 s $$ is (assume that figure shows vertical plane of the cube).
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$$6\ N $$
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$$10\ N $$
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$$15\ N $$
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$$20\ N $$
Explanation
$$r = 4 t^{2} \hat{i} - \dfrac{2}{3} t^{3} \hat{j} $$
$$V = \dfrac{dr}{dt} = 4 \times 2 t - \dfrac{3 \times 2}{3} \hat{j} $$
$$a = \dfrac{dv}{dt} = 8i - 4t \hat{j} $$
$$ F\, to\, f = (mg - 4m) i - (8im)$$
$$|F| = \sqrt{(6)^{2} + (8m)^{2}} $$
$$ = \sqrt{100 m^{2}} = 10 $$N
A man is standing at the center of rotating turntable with his arms outstretched. If he draws his inwards and thereby reduces his moment of inertia by a factor k, the angular speed of the turntable:
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goes down by a factor $$k^{2} $$
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goes down by a factor $$k$$
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remains unaltered
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goes up by a factor $$k$$
Explanation
Concept : If total external torque acting on the system is zero then the momentum remains conserved.
Angular momentum $$ = I \omega $$
$$I$$ = Moment of Inertia
$$\omega $$ = Angular speed
$$I_{1} \omega_{1} = I_{2} \omega_{2} $$
Moment of Inertia is reduced by a factor of $$k$$
$$I \omega = \dfrac{I}{k} \times \omega_{2} $$
$$ \omega_{2} = \dfrac{\omega}{k} $$
Consider the two statements related to circular motion in usual notations.
A. In uniform circular motion, $$\vec {\omega}, \vec {v}$$ and $$\vec {a}$$ are always mutually perpendicular
B. In non-uniform circular motion, $$\vec {\omega}, \vec {v}$$ and $$\vec {a}$$ are always mutually perpendicular.
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Both A and B are true
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Both A and B are false
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A is true but B is false
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A is false but B is true
Explanation
$$\vec {V} = \vec {r}\times \vec {\omega}$$
In uniform circular motion
$$v = constant$$
$$a_{t} = 0$$ (tangential acceleration)
$$a_{c} =$$ Not zero
So,
But In Non-uniform motion.
Velocity, $$\omega$$ are perpendicular but acceleration will be perpendicular because tangential acceleration will be there.
Which of the following temperature will read the same value on celsius and Fahrenheit scales.
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$$-40^{\circ}$$
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$$+40^{\circ}$$
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$$-80^{\circ}$$
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$$-20^{\circ}$$
Explanation
We know that
$$\dfrac {C - O}{100} = \dfrac {F - 32}{180}$$ (When both temperature are equal)
$$\dfrac {C}{100} = \dfrac {C - 32}{180}$$
$$180C = 100C - 3200$$
$$180C = 100C = -3200$$
$$80C = -3200$$
$$\therefore C = \dfrac {-3200}{80} = -40^{\circ}C$$
$$F = -40^{\circ}F$$.
If a particle is thrown upward with speed $$40\,m/s,$$ then the distance travelled by particle in last second of its upward journey is (g is acceleration due to gravity )
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$$g$$
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$$\dfrac{g}{2}$$
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$$3g$$
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$$\dfrac{g}{4}$$
Explanation
Let the horizontal distance be H and
let the distance traveled in the last second be d
$$H = \dfrac{u^2}{2g}$$
$$H = \dfrac{1600}{19.6}$$
$$H = 81.63\,m$$
Now time taken by the body to reach a height of $$81.36\,m = \sqrt{\dfrac{2H}{g}} = 4.08\,s$$
Now from equation of motion, we have
$$s = ut + \dfrac{1}{2}gt^2$$
$$H - d = (40)(t - 1) - \dfrac{1}{2} g(t - 1)^2$$
$$81.63 - d = (40) (4.08 - 1) - \dfrac{1}{2} (9.8) (4.08 - 1)^2$$
$$d = 4.85\,m$$
So, the correct answer is $$4.9\,m$$
A stone thrown down with a seed u takes a time $$t_1$$ to reach the ground while another stone, thrown upwards from the same point with the same speed, takes time $$t_2.$$ The maximum height the second stone reaches from the ground is
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$$\dfrac{1}{2} gt_1t_2$$
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$$\dfrac{g}{8} (t_1 + t_2)^2$$
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$$\dfrac{g}{8} (t_1 - t_2)^2$$
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$$\dfrac{1}{2}gt_2^2$$
A toy completes one round of a square track of side $$2\ m $$ in $$40\ s $$. What will be the displacement at the end of $$3$$ min ?
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$$52\ m$$
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zero
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$$4 \sqrt{2} $$
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$$2 \sqrt{2} $$
Explanation
$$2\ m $$ he travels in $$10$$ sec ($$ 40/4$$ one side)
$$3$$ min $$ \rightarrow 180$$ sec
$$ = 40 + 40 + 40 + 40 + 20 $$
$$ \downarrow $$
Full cycles
Zero displacement
Displacement = AD = $$ 2 \sqrt{2} $$
A stone is dropped from a tower of height $$200\ m $$ and at the same time another stone is projected vertically up at $$50\ m/s.$$ The height at which they meet ?
$$[g = 10\ m/s^{2}] $$
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$$5\ m $$
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$$100\ m$$
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$$120\ m $$
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$$125\ m $$
Explanation
Let meet at $$x$$
$$200 - x = \dfrac{1}{2} gt^{2} $$
For stone upward
$$x = 80t - \dfrac{1}{2} gt^{2} $$
$$x = 50 t - 200 tx $$
$$200 = 50t$$
$$4 = t $$
$$x = 50 \times 4 - \dfrac{1}{2} \times 10 \times 16 $$
$$ = 200 - 80 $$
$$ = 120 $$
The distance travelled by an object $$s$$ is given by
$$s = at^{2} + \dfrac{bt^{3}}{(a + c^{2})} $$, where $$t$$ is time. The dimensions of $$b$$ and $$c$$ respectively are
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$$[L^{2} T^{-5}] \left [L^{\dfrac{1}{2}} T^{-1} \right ]$$
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$$ [L^{3/2} T^{-3}] . \left [L^{\dfrac{1}{2}} T^{-1} \right] $$
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$$[L^{2} T^{-3}] , [L^{-1} T^{-1/2}] $$
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$$[L^{2/3} T^{\dfrac{1}{4}}] [L^{2} T] $$
Explanation
$$s = at^{2} + \dfrac{bt^{3}}{a + c^{2}} $$
$$at^{2} = L $$
a$$a = LT^{-2} $$
$$c^{2} = a $$
$$c = \sqrt{a} $$
$$\boxed{c = L^{1/2} T^{-1}} $$
$$ \dfrac{b}{a + c^2} t^{3} = L $$
$$bt^{3} = LL T^{-2} $$
$$b = \dfrac{L^2 T^{-2}}{T^{3}} $$
$$ b = L^{2} T^{-5} $$
Two persons standing on a floating boat run in succession along its length with a speed $$4.2$$ relative to the boat and dive off from the end. The mass of each man is $$80\,kg$$ and that of boat is $$4\,kg.$$ If the boat was initially at rest, find the final velocity of the boat. Neglect friction:
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$$0.6\,m/s$$
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$$0.7\,m/s$$
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$$0.1\,m/s$$
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$$2.3\,m/s$$
Explanation
Let u be the velocity of the boat after the first man has jumped off. Then the velocity of the first man relative to the ground is $$(4.2 + u)m/s.$$
Since the momentum of the first man is equal and opposite to the momentum of boat and second man, we have
$$80(4.2 + u) = (400 + 80)u$$
$$\Rightarrow u = 0.84\,m/s$$
After the first man has jumped off, the velocity of the boat with second man in it is $$0.84\,m/s$$
$$\Rightarrow$$ momentum of boat with the second man = $$(400 + 80)0.84$$
$$= 403.2 \,kg-m/s$$
Let the velocity of the boat be v m/s after the second man has jumped off.
Momentum of boat = $$400v$$
Momentum of the second man = $$80 \times$$ velocity of man relative to ground
$$= 80(4.2 + v)$$
Momentum of boat and momentum of second man are in opposite direction
$$\Rightarrow$$ net momentum of boat and second man = $$400v - 80(4.2 + v)$$
By conservation of momentum,
$$400v - 80(4.2 + v) = 403.2$$
$$\Rightarrow v = 2.30 \,m/s$$
The least count of stop watch is $$\dfrac{1}{5}$$ second.The time of $$20$$ oscillations of a pendulum is measured to be $$25$$ seconds. The maximum percentage error in the measurement of time will be
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$$0.1\%$$
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$$0.8\%$$
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$$1.8\%$$
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$$8\%$$
The position co-ordinates of an object ae given by $$ x = 2t^{2} , = t^{2} - 4t , z = 3t - 5 $$ the average velocity of the particle during the interval 0 - 1 is
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1010 unit
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12 10 unit
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$$\sqrt{22} unit $$
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8 unit
A researcher found petrified dinosaur faces. Which one of the following is unlikely to be found in this fossil?
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Decayed conifer wood
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Bamboo
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Cyead
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Gian tfern
A particle moving in a straight line has initial velocity u. its accelerations, a is given as $$ a = -k \sqrt{v}$$ where v is the instantaneous velocity and K is a positive constant. the time after which its velocity becomes $$\frac{u}{2}$$ is
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$$ \frac{2\sqrt{u}}{k}$$
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$$ \frac{\sqrt{u}}{k}(\sqrt2-1)$$
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$$ \frac{\sqrt{u}}{k}(2-\sqrt2)$$
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$$ \frac{\sqrt{u}}{2k}(2-\sqrt2)$$
Explanation
We know that
u = -k√v = dv/dt
$$ V_f = u/2$$
$$\implies \vec a = dv/dt = -k√v$$
dv = -k√v ×dt
$$ \displaystyle \int^{u/2}_{u} dv/√v = \int^t_0 -k dt$$
On integration we get
$$\implies t = \dfrac{√u}{k}[2-√2]$$
Hence the correct option s C.
A rod (mass $$m$$, length $$L$$) is hinged at one end, kept in a vertical plane as shown. If a block of mass $$2m$$ moving with velocity of $$v_{0}$$ hits the lower end of the rod elastically. Find the angular velocity of rod just after collision.
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$$6V_{0}/5L$$
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$$3V_{0}/5L$$
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$$6V_{0}/7L$$
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$$12V_{0}/7L$$
A car is moving on a straight road with constant acceleration. The initial velocity of the car is $$5\ m/s$$ and after $$10\ seconds$$ its velocity becomes $$25\ m/s$$. Find Distance travelled by the car in first $$10\ seconds$$.
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$$150\ m$$
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$$160\ m$$
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$$120\ m$$
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$$110\ m$$
Explanation
$$V = u + at$$
$$25 = 5 + a \times 10$$
$$a = \dfrac {20}{10} = 2\ m/s^{2}$$
$$v^{2} - u^{2} = 20s$$
$$(25)^{2} - (5)^{2} = 2\times 2\times 5$$
$$20\times 30 = 4\times 5$$
$$s = 150\ m$$.
A square plate having mass m and side length 'a', its moment of inertia along the axis passing through its diagonal is-
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$$\dfrac{ma^2}{6}$$
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$$\dfrac{ma^2}{12}$$
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$$\dfrac{5ma^2}{12}$$
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$$\dfrac{ma^2}{2}$$
A projectile is thrown with a velocity of $$10\ ms^{-1}$$ at an angle of $$60^{\circ}$$ with horizontal. The interval between the moments when speed is $$\sqrt {5g} m/s$$ is (Take, $$g = 10\ ms^{-2})$$.
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$$1\ s$$
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$$3\ s$$
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$$2\ s$$
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$$4\ s$$
Explanation
In a new system of units, the unit of mass equals $$\alpha$$ kg, the unit of length equals $$\beta$$ m and unit of time equals $$\gamma$$ second. The value of 1 newton in this new system of units is
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$$\dfrac{\alpha\beta}{\gamma^2}$$
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$$\dfrac{\alpha}{\beta\gamma^2}$$
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$$\dfrac{\gamma}{\alpha\beta^2}$$
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$$\dfrac{\gamma^2}{\alpha\beta}$$
Let us consider a system of units in which mass and angular momentum are dimensionless. If the length has dimension of L, then the dimension of power is
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$$L^{-2}$$
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$$L^{-4}$$
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$$L^{-6}$$
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$$L^{3}$$
Explanation
$$[M] = mass = M^0L^0T^0$$
$$J(R)$$ = angular momentum
$$e = mvR = \dfrac{mL^2}{T}$$
$$m \rightarrow$$ Dimensionless
$$T = L^{2}$$
L is also Dimensionless
Now $$P = mv = \dfrac{mL}{T}$$
m is dimensionless
$$[P] = L^{-1}$$
Power = $$\dfrac{mL^2}{T^2 . T} = \dfrac{L^2}{L^2T^2L} = L^{-4}$$
A projectile is fired from level ground at an angle $$\theta$$ above the horizontal. The elevation angle $$\phi$$ of the highest point as seen from the launch point is related to $$\theta$$ by the relation:
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$$\tan \phi=\dfrac{1}{4}\tan \theta$$
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$$\tan \phi=\tan \theta$$
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$$\tan \phi=\dfrac{1}{2}\tan \theta$$
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$$\tan \phi=2\tan \theta$$
Explanation
Now $$\tan \phi=\dfrac{2h}{R}$$
$$h=\dfrac{u^{2}\sin^{2}\theta}{2g}$$
$$R=\dfrac{u^{2}\sin 2\theta}{g}$$
So $$\tan \phi=2\times \dfrac{u^{2}\sin^{2}\theta}{2g}\times \dfrac{2}{u^{2}\sin 2\theta}$$
$$=\dfrac{\sin^{2}\theta}{2\sin\theta\cos \theta}$$
$$\tan \phi=\dfrac{1}{2}\tan \theta$$
A circular loop having centre O formed by using a thin wire of length L and uniform linear mass density $$\lambda$$ as shown in figure. the M.I. of the given loop about the axis 'XX' is
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$$\dfrac{\lambda L^3}{8^2}$$
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$$\dfrac{\lambda L^3}{16^2}$$
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$$\dfrac{5 \lambda L^3}{16^2}$$
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$$\dfrac{3 \lambda L^3}{8^2}$$
A force $$\vec{F}=(3 \vec{i}+4 \vec{j}) \mathrm{N}$$ acts on a particle moving in $$x-y$$ plane. Starting from the origin, the particle first goes along $$x$$ -axis to the point $$(4,0)m$$ and then parallel to the $$y$$ -axis to the point $$(4,3)$$ $$\mathrm{m}$$. The total work done by the force on the particle is
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$$+12\, J$$
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$$-6\, J$$
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$$+24\, J$$
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$$-12\, J$$
Explanation
$$Work=\vec{F} \cdot \vec{s}$$
$$\begin{aligned} W_{(4,0)} &=(3 \hat{i}+4 \hat{j}) \cdot(4 \hat{i}+0\hat{k}) \\ &=12J \end{aligned}$$
$$\begin{aligned} W_{(4,0) \rightarrow (4,3)} &=(3 \hat{i}+4 \hat{j}) \cdot(0 \hat{i}+3\hat{j}) \\ &=12J \end{aligned}$$
$$W_{Total}=24J$$
A force of 100 N need to be applied parallel to a smooth! inclined plane just to hold a body on it. The angle inclination of the inclined plane is $$30^{0}$$. How much horizontal force need to be applied to do the same?
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50 N
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87 N
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100 N
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115 N
Explanation
$$mg \sin 30 = 100$$
$$mg/2 = 100$$
$$mg = 200 \space N$$
$$ma \cos \theta = mg \sin \theta$$
$$ma \cos \theta = 100$$
$$ma \sqrt{3}/2 = 100$$
$$\Rightarrow ma = 200/\sqrt{3} \approx 115 \space N $$
In a rectangle ABCD (BC = 2AB). The moment of inertia is maximum along axis through
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BC
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AB
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HF
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EG
Explanation
perpendicular distance is maximum when the axis of rotation passes through AB, hence Moment of inertia about AB is maximum.
Position of a moving particles is given by $$x=4(1-e^{-2t})$$. Initial velocity of the particle is
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$$8\ m/s$$
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$$4\ m/s$$
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$$10\ m/s$$
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$$Zero$$
Explanation
$$x=4(1-e^{-x4}$$
We know that
$$v=\dfrac{dx}{dt}=\dfrac{d(y)}{dt}-y\dfrac{d(e^{-2t})}{dt}=0-4(e^{-2t})x(-2)$$
$$V=4\times 2e^{-2t}$$
$$V$$ at $$t=0=4\times 2e^{-2\times 0}=2\times 4=8\ m/s$$
Two carts of masses 200 kg and 300 kg on horizontal rails are pushed apart suppose the cofficient of friction between the carts and the rails are same. If the 200 kg cart travels a distance of 36 m and stops, then the distance travelled by the cart weighing 300 kg is
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$$32 m$$
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$$24 m$$
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$$16 m$$
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$$12 m$$
A long capillary tube with both ends open is filled with water and set in a vertical position. The radius of the capillary is $$0.1\ cm$$ and surface tension of water is $$70$$ dynes/ cm. The length of the water column remaining in the capillary tube will be approximately.
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$$1.5\ cm$$
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$$2.9\ cm$$
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$$3.6\ cm$$
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$$5.8\ cm$$
Explanation
Weight of fliuid = Surface tension force at both end
$$\pi r^2 d_0gh =2F=4\pi r\alpha$$
$$h=\dfrac {4\alpha}{d_0 gr}$$
here $$\alpha$$ is surface position
$$h=3\ cm$$
$$3.14\times (0.1)^2 \times 10^{-4}\times 100\times 10\times h$$
$$=4\times 3.14\times 0.1\times 10^{-2}x$$
$$h=\dfrac {4\times 70\times 10^{-5}/ 10^{-2}}{1000\times 10\times 0.1\times 10^{-2}}$$
$$=\dfrac {280\times 10^{-5}}{10^{-1}}$$
$$=280\times 10^{-4}m$$
$$=0.28\times 10^{-2}m$$
$$=0.28\ cm$$
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