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CBSE Questions for Class 8 General Knowledge Basic Science Quiz 6 - MCQExams.com
CBSE
Class 8 General Knowledge
Basic Science
Quiz 6
The resultant of two forces at right angles is $$13N$$. The minimum resultant of the two forces is $$7N$$. The forces are
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0%
$$20\,N, 6\,N$$
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$$10\,N, 20\,N$$
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$$5\, N, 12\, N$$
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$$8\, N, 15\, N$$
A body is thrown up with a velocity $$29.23ms!$$ $$V$$ distance travelled in last second of upward motion is
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$$2.3m$$
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$$6m$$
0%
$$9.8m$$
0%
$$4.9m$$
Explanation
Total time of motion:
$$v=u+at$$
$$x_0=29.23-9.8\times 6$$
$$xt=2.98\ sec$$
distance total
$$\Rightarrow v^2=u^2+2as$$
$$\Rightarrow 0=(29.23)^2=2\times 9.8 \times s$$
$$\Rightarrow s=43.59\ m$$
Now in $$1.96\ sec$$
$$s=29.3\times 1.98-\dfrac{1}{2}\times 9.8 (1.98)^2$$
$$s=36.6654$$
in last $$s= 43.59-38.66$$
$$=4.93m$$
What is the angle between $$\vec{i} +\vec{j}$$ and $$\vec{i}$$?
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$$0$$
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$$\pi/6$$
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$$\pi/3$$
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$$\pi/4$$
Explanation
We know that
$$\cos \theta=\dfrac{|\vec{i} +\vec{j}.\vec{i}|}{|\vec{i} +\vec{j}||\vec{i}|}$$
$$\cos \theta=\dfrac{1}{\sqrt{2}.1}=\dfrac{1}{\sqrt{2}}$$
$$\theta=45^\circ$$
The refractive index of glass is measured and values are $$1.49,1.50,1.52,1.54$$ and $$1.48$$. Percentage error in the measurement is
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$$0.0132$$%
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$$1.32$$%
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$$0.56$$%
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$$0.156$$%
Explanation
Now mean absolute error $$=({y}_{1}+{y}_{2}+{y}_{3}+{y}_{4}+{ty}_{5})/5$$
$$(0.01+0.00+0.02+0.04+0.002)/5=0.09/5=0.018$$
relative error $$=$$ mean absolute error/ mean value
$$=0.018/1.50=0.012$$
Now % error in the measurement $$=100\times relative error=100\times 0.012=1.2$$%
Which temperature is same on centigrade as well as on Fahrenheit scale?
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$$40$$
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$$-40$$
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Both
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None
Explanation
$$F=\left(\dfrac{9}{5}\right)C+32$$
as $$F=C$$
$$C=\left(\dfrac{9}{5}\right)C+32$$
$$C-\dfrac{9}{5}C=32$$
$$\dfrac{-4C}{5}=32$$
$$C=-40$$
Initial velocity of a particle moving along straight line with constant acceleration is $$(20 \pm 2) \ m/s$$, if its acceleration is $$a=(5 \pm 0.1) \ m/s^2$$, then velocity of particle with error, after time $$t=(10 \pm 1) \ s$$, is
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$$(70 \pm 3.1) \ m/s$$
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$$(70 \pm 2) \ m/s$$
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$$(70 \pm 4) \ m/s$$
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$$(70 \pm 8) \ m/s$$
A ballet dancer spins about vertical axis at $$120$$ rpm, with her arms out stretched, With her arms folded, the M.I. about the same axis of rotation decrease by $$40\%$$. What will be the new rate of revolution ?
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$$500$$ rpm
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$$300$$ rpm
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$$100$$ rpm
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$$200$$ rpm
Explanation
$$w = 120$$ decrease by $$40\%$$
$$I' = I - I \cdot \dfrac{4}{10} \Rightarrow I'= \dfrac{3}{5}I$$
now apply energy conservation
$$\dfrac{3}{5}I \cdot w_0 = I \cdot 120$$
$$w_0 = 120 \times \dfrac{5}{3} = 200 \,r.p.m.$$
A car travelling at a speed of 30 kilometer per hour is brought to a halt in 8 metres by applying brakes. If the same car is travelling at 60 km per hour, it can be brought to a half with same braking power in
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$$8$$ metres
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$$16$$ metres
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$$24$$ metres
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$$32$$ metres
Explanation
Applying kinetic energy theorem
$$\Delta KE = \Delta w$$$$\frac{1}{2}mv^2 = F \cdot s$$
$$v^2 \alpha s$$
$$\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{s_1}{S_2}$$
$$v_2 = 2v_1$$
$$= \frac{1}{4} = \frac{s_1}{S_2}$$
$$s_2 = 4s_1$$
$$= 4 \times 8$$
$$= 32$$
Hence (A) option is correct answer
An elevator initially at rest supported by a cable passes over a pulley of radius $$ 0.4m.$$ It is accelerated upwards with $$ 0.6 m/s^{2} $$.Find number of rotations made by pulley during $$ 10s. $$
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23.88
0%
11.94
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5.97
0%
8.96
Explanation
$$ S =\frac{1}{2} at^{2} $$
$$ = \frac{1}{2} \times 0.6 \times 10 \times 10 $$
$$ = 30m $$
$$ \therefore n \times 2\pi r = 30 $$
$$ n = \frac{30}{2 \times {22}{7} \times {4}{10}} $$
$$ = 11.94 $$
A table fan rotating at a speed of $$2400$$ rpm is switched off and the resulting variation of the rpm with time is shown in the figure. The total number of revolutions of the fan before it come to rest is:
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$$420$$
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$$280$$
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$$190$$
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$$16800$$
Explanation
Area under the curve will be the no of revolution
Area $$= \dfrac{1}{2} \times (40 - 10) \times 8 + (10 \times 8) \times \dfrac{1}{2} \times 10 \times (24 - 8) = 120 + 80 + 80 = 280$$
A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleraction due to gravity, then its range is:
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$$\dfrac{4v^2}{5g}$$
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$$\dfrac{4g}{5v^2}$$
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$$\dfrac{4v^3}{5g^2}$$
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$$\dfrac{4v}{5g^2}$$
A body slides down a smooth inclined plane of base $$x$$ meter. Find the angle of the plane so that the time taken by the body to reach the bottom of the plane is minimum
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$$60^{\circ}$$
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$$30^{\circ}$$
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$$45^{\circ}$$
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$$53^{\circ}$$
Explanation
$$bass \rightarrow x meter$$
$$cos \theta = \dfrac{x}{d}$$
$$d = (\dfrac{x}{cos\theta })$$
$$d \rightarrow $$$$S = ut + \dfrac{1}{2}a t^2$$$$d = 0 + \dfrac{1}{2}g sin\theta t^2$$
$$\sqrt{\dfrac{2d}{gsin\theta } = t}$$
$$t \rightarrow $$ time taken $$\dfrac{dt}{d\theta } = 0$$
$$\dfrac{dt}{d\theta } = \sqrt{\dfrac{2d}{9}} \times \dfrac{1}{\sqrt{sin \theta }}$$
$$= (sin\theta )^{-1/2}$$
$$= -\dfrac{1}{2} (sin \theta )^{-3/2} \times cos \theta = 0$$
$$\dfrac{dx}{d\theta } = 0$$
$$\theta= 45$$
A force $$\overrightarrow{F}= (x \hat{i} + 2y \hat{j} ) N$$ is applied on an object of mass $$10 \ kg$$. Force displaces the object from position $$A(1,0) \ m$$ to position $$B(3,3) \ m$$ then the work done by the force is ($$x$$ and $$y$$ are in meter)
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$$8 \ J$$
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$$5 \ J$$
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$$13 \ J$$
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$$16 \ J$$
Explanation
$$W= \displaystyle\int \overrightarrow{F}. d \overrightarrow{x}$$
$$= \displaystyle\int^3_1 x \ dx + \displaystyle\int^3_0 2y \ dy$$
$$= \left[\dfrac{x^2}{2} \right]^3_1 + \left[ \dfrac{2y^2}{2} \right]^3_0$$
$$= \dfrac{3^2}{2} - \dfrac12 + 3^2$$
$$= 4+9$$
$$=13 \ J$$
With what velocity must a body be thrown from Earth's surface so that it may reach a height $$4$$R above the Earth's surface. (Radius of Earth Re= $$6400km,g=9.8m/s^2$$)
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$$20km$$
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$$10km$$
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$$15km/s$$
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$$22km/s$$
Explanation
According to question
We have $$v^2=\frac{8}{5}*g*Re$$
$$v^2=\frac{8}{5}*10*6400$$
$$v=4*80$$
$$v=320m/s$$
$$v^2=\frac{8}{5}*9.8*6400000$$
$$v=\sqrt{\frac{8*9.8*6400000}{5}}$$
$$v=10017.58m/s$$
$$v=10.017km/s$$
A man of $$50kg$$ mass is standing in a gravity free space at a height of $$10m$$ above the floor. He throws a stone of $$0.5kg$$ mass downwards with a speed of $$2m/s$$. When the stone reaches the floor, the distance of the man above the floor will be
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$$20m$$
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$$9.9m$$
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$$0.1m$$
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$$10m$$
Explanation
We know that
$$X_{com}=0$$
$$\frac{M_{mass}X_{mass}+M_{stone}X_{stone}}{M_{stone}+M_{mass}}=0$$
$$X_{mass}=\frac{-M_{stone}X_{stone}}{M_{mass}}$$
$$-\frac{0.5*10}{50}=0.1m$$
A stone falls freely under gravity. It covers distance $$h_1$$ and $$h_2$$ in the first $$2$$ seconds and the next $$2$$ seconds respectively. The relation between $$h_1$$ and $$h_2$$ is
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$$h_1=h_2$$
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$$h_1=\frac{1}{3}h_2$$
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$$h_1=3h_2$$
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$$h_2=4h_1$$
Explanation
We know that
$$h_1=ut + \frac{1}{2}*g*(t)^2$$
Therefore we have
$$h_1=\frac{1}{2}*g*(2)^2$$
$$=\frac{4g}{2}=2g$$
$$v_2=10*2*20$$
$$h_2=20*2+\frac{1}{2}*g*2^2$$
$$=40+20=60$$
$$\frac{h_1}{h_2}=\frac{29}{60}=\frac{1}{3}$$
$$h_1=\frac{h_2}{3}$$
The resultant of two forces $$10N$$ and $$15N$$ acting $$+x$$ and $$-x$$ axes resp.
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$$25N$$ along $$+x$$ axis
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$$25N$$ along $$-x$$ axis
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$$5N$$ along $$+x$$ axis
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$$5N$$ along $$-x$$ axis.
In a vernier callipers , one scale division in x cm , and n division of the vernier scale coincide with (n -1) divisions of the main scale . The least count (in cm ) of the callipers is
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$$ \left ( \dfrac{n -1}{n} \right )x $$
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$$ \dfrac{nx}{(n -1)} $$
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$$ \dfrac{x}{n} $$
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$$ \dfrac{x}{(n- 1)} $$
For a particle having a uniform circular motion, which of the following constant?
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Speed
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Acceleration
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Velocity
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Displacement
Explanation
In a uniform circular motion speed is constant. Other are vector quality that is changing.
One stick leans on another as shown in figure. A right angle is formed where they meet, and the right stick makes an angle 0 with the horizontal. The left stick extends infinitesimally beyond the end of the right stick. The coefficient of friction between two sticks is $$\mu$$. The sticks have the same mass density per unit length and are both hinged at the ground. What is the minimum value of $$\tan \theta$$ for which the sticks do not fall? (Take $$\mu = 0.04$$)
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0%
5
0%
3
0%
8
0%
12
Explanation
According to given conditions
We have
$$(\tan \theta)^2 \geq \frac{1}{\mu}$$
$$\tan \theta \geq \frac{1}{\sqrt{\mu}}$$
$$ \geq \frac{1}{\sqrt{0.04}}$$
$$ = 5$$
A block on an inclined plane, as shown above, has four forces acting on it.
$$\textbf{W:} $$ the force of gravity acting downward (weight)
$$\textbf{N:} $$ a force normal to the surface of the plane
$$\textbf{f:} $$ a friction force acting along the plane
$$\textbf{F:} $$ an external applied force acting along the plane.
The following choices illustrate various possibilities when these forces are added, depending on the magnitude and directions of $$ f $$ and $$ F $$.
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0%
0%
0%
0%
A physical quantity Q is found to depend on observable x, y and z obeying relation $$ Q= \frac {x^3 y^2}{z} $$ the percentage error in the measurements of x, y and z are $$ 1$$ %, $$ 2$$ % and $$ 4$$ % respectively. what is percentage error is the quantity Q.
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$$ 11 $$ %
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$$ 4 $$ %
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$$ 1 $$ %
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$$ 3 $$ %
Explanation
$$ Q = \dfrac {x^3 y^2}{ z}$$
$$ \dfrac { \Delta Q}{Q} \times 100 = 3 \dfrac { \Delta x}x \times 100 + 3 \times \dfrac { \Delta y}{y} \times 100 + \dfrac { \Delta Z}{z} \times 100 $$
$$ \Delta Q= 3 \times (1) +2 (2) + \Delta $$
$$ \Delta Q= 3+ 4 + 4 = 11 $$ %
The dimension of heat capacity are
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$$ \left [ ML^{-2} T^{-2}K^{-1} \right ] $$
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$$ \left [ ML^{2} T^{-2}K^{-1} \right ] $$
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$$ \left [ M^{-1}L^{2} T^{-2}K^{-1} \right ] $$
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$$\left [ MLT^{-2} K \right ] $$
Explanation
Answer
(Heat capacity) = $$ = \left [ \dfrac{Energy}{Temeperature} \right ] $$
$$ = \left [ \dfrac{F \times l}{k} \right ] = \dfrac{MLT^{-2} \times L}{K} $$ $$ML^{2}T^{-2}K^{-1} $$
For a particle in uniform circular motion the acceleration $$\vec{a}$$ at a point $$(R,\theta)$$ on the circle of radius $$R$$ is (Here $$\theta$$ is measured from x-axis)
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$$-\frac{v^2}{R}\cos\theta\hat{i}+\frac{v^2}{R}\sin\theta\hat{j}$$
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$$-\frac{v^2}{R}\sin\theta\hat{i}+\frac{v^2}{R}\cos\theta\hat{j}$$
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$$-\frac{v^2}{R}\cos\theta\hat{i}-\frac{v^2}{R}\sin\theta\hat{j}$$
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$$\frac{v^2}{R}\hat{i}+\frac{v^2}{R}\hat{j}$$
Explanation
$$a_c=\frac{v^2}{R}$$ (towards the centre)
Component in x-direction = $$-\frac{v^2}{R}\cos\theta$$
Component in y-direction = $$-\frac{v^2}{R}\sin\theta$$
$$\vec{a_c}=-\frac{v^2}{R}\cos\theta\hat{i}-\frac{v^2}{R}\sin\theta\hat{j}$$
A clerk starts from his house with a speed of $$ 2 kmh^{-1} $$ and reaches the office 3 min late .Next day the increase speed $$ 1 kmh^{-1}$$ and reaches the office 3 min earliar .Find the distance between his house and office .
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$$7 km$$
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$$ 600m $$
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$$ 700m $$
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$$8000m $$
Explanation
Ratio of speed $$=2:3$$
hence Ratio of line $$=3:2$$ (inverse of speed)
gap in time $$=6\ min$$
New $$\Rightarrow $$ gap in Ratio $$=3-2=1=6\ min$$
$$3=18\ min$$
$$2=12\ min$$
distance between o`ffice & how $$=\dfrac {12}{60}\times 3=600\ m$$
Three blocks of masses $$4kg,6kg$$ and $$8kg$$ are hanging over a fixed pulley shown. The tension in the string connecting $$4kg$$ and $$6kg$$ block is: $$(g=10m/s^2)$$
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$$4N$$
0%
$$6N$$
0%
$$\frac{320}{9}N$$
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$$\frac{40}{9}N$$
Explanation
By figure we have
Onbalaming the forces we get
$$T_1-8g=8a$$
$$T_2+6g-T_1=6a$$
$$4g-T_2=4a$$
$$2g=18a$$
$$a=\frac{g}{9}$$
$$4g-T_2=\frac{4g}{9}$$
$$T_2=4g-\frac{4g}{9}$$
$$=360-\frac{40}{9}$$
$$=\frac{320}{9}N$$
Two solid spheres of same mass are having densities $$d_1$$ and $$d_2(d_1<d_2)$$. Their moments of inertia about diameter are $$I_1$$ and $$I_2$$ respectively. Then,
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$$I_1 > I_2$$
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$$I_1 < I_2$$
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$$I_1 < I_2$$
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$$I_1d_2=I_2d_2$$
Explanation
MI of the solid sphere about diameter = $$\frac{2}{3}MR^2$$
$$M=$$ density * volume
$$M_1=d_1(\frac{4}{3}\pi R^3)$$
$$M_2=d_2(\frac{4}{3}\pi R^3)$$
$$I_1=\frac{3}{5}M_1R_1^2$$
$$I_2=\frac{3}{5}M_2R_2^{2}$$
$$\frac{I_1}{I_2}=\frac{\frac{3}{5}d_1(\frac{4}{3}\pi R^3)R^2}{\frac{3}{5}d_2(\frac{4}{3}\pi R^3)R^2}$$
$$\frac{I_1}{I_2}=\frac{d_1}{d_2}$$
as $$(d_1 < d_2)$$
So, $$M_1=M_2=M$$
$$(\frac{4}{3}\pi R_1^3)d_1=M=(\frac{4}{3}\pi R_{2}^3)d_2$$
$$(\frac{R_1}{R_2})^3=(\frac{d_2}{d_1})$$
$$\frac{R_1}{R_2}=(\frac{d_2}{d_1})^{1/3}$$
$$\frac{I_1}{I_2}=\frac{\frac{3}{5}MR_1^2}{\frac{3}{5}MR_2^2}=(\frac{R_1}{R_2})^2$$
$$\frac{I_1}{I_2}=(\frac{d_2}{d_1})^{2/3}$$
as $$d_2 > d_1$$
so, $$\frac{d_2}{d_1} > 1$$
So, $$(\frac{d_2}{d_1})^{2/3} > 1$$
$$\frac{I_1}{I_2} > 1$$
$$I_1 > I_2$$
Maximum percentage error in the measurement of 25.00 m is
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0%
0.01%
0%
0.03%
0%
0.04%
0%
0.02%
Explanation
L = 25.00 meter
maximum error = 1 cm
$$ \dfrac{1cm}{25m}\times{100} $$
= $$ {10}_{-2}\times{4} % $$
= 0.04%
Blocks $$A$$ and $$B$$ have masses of $$2kg$$ and $$3kg$$ respectively. The ground is smooth. $$P$$ is an external force of $$10N$$. The force exerted by $$B$$ on $$A$$ is
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0%
$$4N$$
0%
$$6N$$
0%
$$8N$$
0%
$$10N$$
Explanation
Acceleration is given by $$a=\frac{F}{m_1+m_2}=\frac{10}{5}=2$$
$$F=m_{2}a$$
Force exerted by B on A is given by
$$=2*2=4$$N
Glucose contains
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0%
$$20\%$$ carbon by mass
0%
$$40\%$$ carbon by mass
0%
$$50\%$$ carbon by mass
0%
$$60\%$$ carbon by mass
Explanation
$$M.W_{C_6 H_{2}O_6}=12\times 6+12\times 1+6\times 16$$
$$=180$$
total mass of carbon $$=12\times 6=72$$
$$\therefore \dfrac {72}{180}\times 100=40\%$$
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