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CBSE Questions for Class 8 General Knowledge Basic Science Quiz 7 - MCQExams.com
CBSE
Class 8 General Knowledge
Basic Science
Quiz 7
Which of the following instrument is the least precise device for measring length?
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Vernier callipers with 50 division of Vernier scale equal to 49 division of main scale and each division of main scale is of 1 mn.
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A screw scale having 100 divisions and each division is of 1 cm.
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A metre scale having 100 divisions and each division is of 1 cm.
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An optical instrument that can resolve upto $$10^{-7} m$$.
Explanation
Most precise device is the one with the minimum least count
1. Vernier Calliper
20 divisions on the sliding scale
L.C= $$\frac{1}{20}mm$$ = 0.05mm
2. Screw Gauge
pitch 1 mm and 100 divisions on the circular scale
L.C= $$\frac{1}{100}mm$$ = 0.01mm
3. meter Scale
pitch 1 mm and 100 divisions on the scale
L.C= $$\frac{1}{100}mm$$ = $$10^{-2}mm$$
4.
optical instrument
within a wavelength of light
Wavelenght of light
$$\approx$$
5
8
9
n
m
Hence option D is correct
A vernier callipers (with least count = 0 .1 mm ) has 20 divisions of the vernier scale .The main scale division are of
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0 .2 mm
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0.5 mm
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1.0 mm
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2.0 mm
Explanation
Answer
L.C = M .S.P - IV .S .D
L.C = 0 .1 mm
0.1 mm = M.S .D - 1.8 V.S .D
M.S.D = 0.1 + 1.9
= 2 mm
The diameter of a disc is $$1m$$. It has a mass of $$20kg$$. It is rotating about its axis with a speed of $$120$$ rotations in one minute. Its angular momentum in Kg $$m^2/s$$ is
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$$3.4$$
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$$31.4$$
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$$41.4$$
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$$43.4$$
Explanation
We know that
$$L=I\omega$$
$$L=\frac{Mr^2}{2}*2\pi n$$
$$=\frac{20*1^2}{2}+\frac{2*3.14*120}{60}$$
$$=31.4$$
M gram of ice at $$0^{\circ} C$$ is mixed with M gram of water at $$10^{\circ} C.$$ The final temperature is
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$$8^{\circ} C$$
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$$6^{\circ} C$$
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$$4^{\circ} C$$
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$$0^{\circ} C$$
Explanation
Let final temperature = T then
Heat required to melt ice and to increase temperature by T
$$H_1 = ML + MC\Delta T$$
$$T = \dfrac{-530}{2} = -265^{\circ} C$$
But it can reach only $$0^{\circ} C.$$
So partly the ice will melt and all the water will be at $$0^{\circ} C.$$
A body takes $$5$$ min to cool from $$80^\circ C$$ to $$70^\circ C.$$ To cool from $$80^\circ C$$ to $$60^\circ C,$$ it will take (Room temperature = $$40^\circ C$$)
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$$5$$ min
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$$10$$ min
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$$12$$ min
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$$14$$ min
Explanation
Newton's law of cooling, we have
$$mc(T_1 - T_2) = k \left [ \dfrac{(T_1 + T_2)}{2} - T_0 \right ]$$ .......(i)
$$\dfrac{mc}{K} = \left [ \dfrac{(T_1 + T_2)}{2} - T_0 \right ] / \left [ \dfrac{(T_1 - T_2)}{t} \right ]$$
$$\dfrac{mc}{K} = \left [ \dfrac{80 + 70)}{2} - 40 \right ] / \left [ \dfrac{(80 - 70)}{5} \right ]$$
Substituting
$$T_{1} = 80$$ & $$T_{2} = 60$$ & $$T_{0} = 40$$
equation (i)
$$mc \dfrac{(80 - 60)}{t} = K \left [ \dfrac{(80 + 60)}{2} - 40 \right ]$$
$$20\dfrac{mc}{t} = k \times 30$$
$$t = \dfrac{20\,mc}{30\,K}$$
$$= \dfrac{20}{30} \times \dfrac{35}{2}$$
$$= \dfrac{70}{6}$$
$$t = 11.66 = 12\,min$$
Choose the correct statements from the following. The range of a projectile depends upon
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The angle of projection
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The acceleration due to gravity
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The magnitude of the velocity of projection
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The mass of the projectile
Explanation
The horizontal displacement of the projectile is called the range of the projectile and depends on the initial velocity of the object. If an object is projected at the same initial speed, but two complementary angles of projection, the range of the projectile will be the same.
Which of the statement are correct ?
(1) An electron $$4p$$ orbital has more energy than one in $$3p$$ orbital
(2) An electron in $$4s$$ orbital is further from the nucleus than one in a $$3s$$ orbital
(3) the angular quantum number $$1$$ is related to orbital shape.
(4) Energy is released if electron moves from $$n=2$$ to $$n=5$$ level.
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$$1,2$$ and $$3$$
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$$1$$ and $$3$$
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$$2$$ and $$4$$
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Only $$4$$
Explanation
$$n+l$$rule
1) $$4p=4+1$$
$$3p=3+1$$
$$4p>3p$$
2) energy of $$4s>3s$$ so $$4s$$ is further
3) correct
4) $$2$$ to $$5$$ energy is absorbed $$x$$
A wedge $$ABC$$ and a block are placed as shown in the figure. The is no friction at any surface. Displacement of wedge $$ABC$$ when $$2\ kg$$ block reaches the point $$B$$:
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$$1.6\ m$$
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$$2\ m$$
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$$1.5\ m$$
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$$1.2\ m$$
Explanation
$$\because$$ No ant. force is present in horizontal direction, $$\Delta x_{cm}=0$$
$$\Rightarrow _1 \Delta x_1=m_2 \Delta x_2$$
$$\Rightarrow 2 \times (y-x)=8x$$
$$\Rightarrow y-x=4x$$
$$\Rightarrow 5x=y$$
Now, from trigo
$$\tan 37^o=\dfrac{6}{y}$$
$$\Rightarrow y=6\ cot 37^o$$
$$\Rightarrow x=\dfrac{6}{5} cot 37^o$$
A hypothetical vernier scale of a travelling microscope has 40 divisions which is equal to 38 main scale divisions if each main scale division is 1.2 mm then minimum error in the measurement of length is
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$$ 0.6 mm $$
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$$ 1.2 mm $$
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$$ 0.06 mm $$
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$$ 0.1 mm $$
Explanation
We know that
$$ 40 VSR = 3 SMS R $$
$$ WSR = \frac { 38}{40} $$
least count $$ = 1 - \frac {38}{40} MSR $$
$$ = \frac {2}{40} = \frac {1}{20} \times 1.2 = 0.06 $$
(i) Joule is the unit of force
(ii) Positively charged ion are called anions
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True
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False
Explanation
(I) $$ \rightarrow $$ Force $$ \rightarrow $$ Newton
Joule is unit of work
(Ii) $$ \rightarrow $$ anions $$ \rightarrow $$ negatively change
A beta particle is a fast-moving electron. Which statement explains how beta particles are emitted from an atom?
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An electron is emitted as a beta particle from an inner electron shell of the atom
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An electron is emitted as a beta particle from an outer electron shell of the atom
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A neutron changes into a proton and a beta particle is emitted from the nucleus
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A proton changes into a neutron and a beta particle is emitted from the nucleus
Explanation
$$\therefore $$ The emission of beta particles is from the nucleus due to the nucleus conversion:
$$n\rightarrow p+\beta ^-+\vec v_{e}$$
A body of mass $$5\ kg$$ moving horizontally at a speed of $$1.5\ ms^{-1}$$. A perpendicular force of $$5\ N$$ acts on it for $$4$$ second distance of the body from the point of application of the force is
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$$25\ m$$
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$$6\ m$$
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$$8\ m$$
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$$10\ m$$
Explanation
$$F=\dfrac {dm}{dt}v$$
$$=\dfrac {50}{1000}\times 500$$
$$=25 N$$
A car moving with speed 30 m/s on circle of radius 500 m. Its speed is increasing at of $$2 m/s^{2} $$. The acceleration of the car is
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$$ 9 .8 m/s ^{2} $$
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$$ 1 .8 m/s ^{2} $$
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$$ 2 m/s ^{2} $$
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$$ 2 .7 m/s ^{2} $$
Explanation
Answer
Tangentical acceleration $$a_t= 2m/sec^{2} $$
Velocity (v) = 30 m /sec
Radius (R) = 500 m
$$ (a_{c}) = \dfrac{v^{2}}{R} = \dfrac{(30)^{2}}{R} $$
$$ = \dfrac{900}{500} = 1.8 m /sec{2} $$
$$ = \sqrt{a_t^{2} + a_c^{2}} $$
$$ = \sqrt{4 + 3. 24} $$
$$ 2 .7 m/sec^{2}$$
$$ a_{net} = \sqrt{4 + 3.24} $$
$$a_{net}= \sqrt{7.24} $$
$$a_{net}= 2. 99 = 2.7 $$
The moment of inertia of a solid sphere , about an axis parallel to its diameter and at a distance of a from it is . , I ' (x) ' which one of the graph represents the variation of I ( x ) with x correctly
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0%
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The velocity - time graph for a particle travelling along a straight line as shown in figure , if the graph is a sine curve , the average velocity during the first 5 s is 5.8 m /ms
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6.37 m /s
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2.37 m/s
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2.37 m/s
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7.5 m/s
The gravitational force in region given by $$ \vec{F} = a y \hat{i} + ax \hat{j} $$. The work done by gravitational force to shift a point ,mass m from ( 0 , 0 , 0) to $$ (x_{0}, y_{0},z_{0}) $$ is
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$$ max_{0}y_{0}z_{0} $$
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$$ max_{0}y_{0} $$
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$$ - max_{0}y_{0} $$
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0
The moment of inertia of a wheel is $$ 100 kg - m ^{2} $$ . At a given instant , its angular velocity is 10 rad /s . after the wheel rotates through an angle of 100 radians the wheel 's angular velocity is 100 rad /s . The torque applied on wheel is
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$$4.95 \times 10^{5} N. m $$
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$$4.95 \times 10^{4} N. m $$
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$$4.95 \times 10^{3} N. m $$
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$$49.5 \times 10^{5} N. m $$
A body is dropped from a balloon moving up with a velocity of $$4\ ms^{-1} $$ when the balloon is at a height of $$120.5\ m $$ from the ground; the height of the body after five seconds from the ground is :
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$$8\ m $$
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$$12\ m $$
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$$18\ m $$
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$$24\ m $$
Explanation
Using second equation of newton
$$S = ut + \dfrac{1}{2} at^{2} $$
$$ = 4 \times 5 + \dfrac{1}{2} \times 9.8 \times (5)^{2} $$
$$ = 20 + \dfrac{1}{2} \times 9.8 \times 25 $$
$$ = 20 + 82.5$$
$$=102.5$$
Hence present height $$ = 120.5 - 102.5$$
$$ = 18\ m $$
The maximum positive displacement $$x$$ is
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$$ 2\sqrt{3}\ m $$
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$$2\ m $$
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$$4\ m $$
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$$\sqrt{2}\ m $$
A square metal frame is made of four identical thin rods each of the mass 6 kg and lengthm. The moment of inertial of the frame is $$ kg - m ^{2} $$ about an axis passing through point O and perpendicular to the plane of the frame is
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4
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8
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12
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1
The unit of expression $$\mu_{0} \varepsilon_{0} $$ are :-
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$$m/s$$
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$$m^{2} /s^{2} $$
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$$s^{2} / m^{2} $$
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$$s/m$$
Explanation
$$C = \dfrac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} $$
$$C$$ - Speed of light
$$ \sqrt{\mu_{0} \varepsilon_{0}} = \dfrac{1}{C}$$
$$ \mu_{0} \varepsilon_{0} = \dfrac{1}{C^2} = \dfrac{1}{\left (\dfrac{m}{s} \right )^{2}} $$
$$ = \dfrac{S^2}{m^2} $$
Consider the situation as shown in the figure. The system is free to move.
The force exerted by the rope on the block is FM/(m + m)
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only if the rope is uniform
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in gravity free space
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only if F > (M + m)g
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in all cases
Figure given below shows the graph of velocity $$v$$ of particle moving along $$x-axis$$ as a function of time $$t$$. Average acceleration during $$t = 1\ s$$ to $$t = 7\ s$$is
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$$1.5 m/s^{2}$$
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$$1 m/s^{2}$$
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$$2 m/s^{2}$$
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$$2.5 m/s^{2}$$
Explanation
$$ a = \frac {dv}{dt} = $$ solpe of v-f
$$ \frac { 20 -0}{2 - 0} = 10 m/s^2 0-2 $$
$$ 2- 6 $$
$$ \frac {30-20}{8-6} = \frac {10}{2} = 5 m/s^2 = 6.8 $$
Average acc$$ \Rightarrow \frac { 10 \times 1+ 0 \times 4 +5 \times 1}{6} $$
$$ =\frac {15}{6} = 2.5 $$
a stationary where starts roatating about its own axis at constant angular acceleration if the wheel completes 50 rotatations made by it in next two seconds is
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$$ 75 $$
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$$ 100 $$
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$$ 125 $$
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$$ 150 $$
Explanation
$$ \theta = cot + \frac {1}{2} \alpha t^2 $$
$$ 100 M = \frac {1}{2} \times (2)^2 $$
$$ \alpha = 50 \pi ral /s^2 $$
$$ \therefore \theta $$ for 2 next / second
$$ = 40 com - 1 com $$
$$ = 300 \pi $$
$$ \therefore $$ no f revolute = $$ \frac { 300 M}{2M} = 150 $$
A metal tape gives correct measurement at $$15^{\circ}C$$. It is used to measure a distance of $$100m$$ at $$45^{\circ}C$$. The error in the measurement, if $$\alpha = 12\times 10^{-6}/1^{\circ}C$$ is
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$$36\ cm$$
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$$36\ m$$
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$$42\ mm$$
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$$36\ mm$$
Explanation
Correct measurement at $$15^{\circ}C$$
$$\% error = \dfrac {\triangle l}{l}\times 100$$
$$= \dfrac {l\alpha \triangle T}{l} \times 100$$
$$\dfrac {\triangle l}{l} \times 100 = 12\times 10^{-6} \times (45 - 15) \times 100$$
$$\triangle l = l\times 12\times 10^{-6} \times 30$$
$$= 100\times 12\times 10^{-6} \times 30$$
$$= 360\times 10^{-6} = 36\ mm$$.
How many ATP molecules are formed from one molecule of acetyl $$CoA$$ in $$TCA$$ cycle
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$$1$$
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$$13$$
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$$24$$
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$$2$$
Two equally charged identical pith balls are suspended by identical massless strings as shown in the adjacent figure if this set up is mercury $$ ( g = 3.7 m/s^2)earth ( g = 9.8 m/s^2) $$ and $$ Jupiter ( g =m/s^2) $$ then angle $$ 2 \theta $$ will be
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maximum on mercury
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Maximum on earth as it ha atmosphere
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maximum on jupiter
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the same on any planet as coulomb force is independent of gravity
Explanation
$$ tan \theta = \frac {F}{ mg} = \frac { \frac { kq^2}{ 4 sin^2 \theta } }{ mg} $$
$$ = \frac { kq^2}{ 4 sin^2 \theta mg } $$
$$ tan \theta \times in^2 \theta = \frac {kq^2}{ 4 m g } $$
$$ \frac { sin^3 \theta }{ cos \theta } = \frac { ka^2}{4 m} ( \frac {1}{g} ) $$
$$ tan^2 \theta = sin \theta \alpha \frac {1}{g} $$
$$ \theta \alpha \frac {1}{g} $$
So $$ \theta $$ is max when g is min so g is on mercury
Two particle A and B are thrown vertically upward simultaneously , with velocity 5 m/s and 15 m/s respectively from the same position the separation between them after 1 s is $$ ( g = 10 m/s^2) $$
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$$ 7.5 m $$
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$$ 5 m $$
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$$ 10 m $$
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$$ Zero $$
Explanation
$$ S_1 - \mu t - l_2 gl^2 $$
$$ = 5 \times 1 - \frac {1}{2} \times 10 l^2 $$
$$ = S - S $$
$$ = OM $$
$$ S_2 = \mu_2 t - t_2 gl^2 $$
$$ = 15 \times 1 - \frac {1}{2} \times 10 $$
$$ = 15 - 5 = 10 m $$
$$ \Delta S = S_2 - S_1 = 10 - 0 = 10 m $$
A particle ( 5 kg) moves upon straight line with constant velocity $$ 2 ms^{-1} $$ along the straight line $$ 2y = 3 x+ 4 $$ the angular momentum of particle about origin in $$ kg m^2s^{-1} $$ will be
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$$ \frac {30}{ \sqrt { 13} } $$
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$$ \frac { 40 }{ \sqrt {13} } $$
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$$ \frac {20}{ \sqrt { 13} } $$
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$$ \frac {10}{ \sqrt {13} } $$
Explanation
Angular momentum
$$ |Rightarrow m ( \overrightarrow {r} \times \overrightarrow {V} ) = m \mu r $$
so velocity using x
direction will be $$ = 2 \times cos \theta $$
$$ \Rightarrow 2 \times \frac {2}{ \sqrt {3}}= \frac {4}{ \sqrt { 13} } m/s $$
Velocity along y direction
$$ \Rightarrow 2 sin \theta = 2 \times \frac {3}{ \sqrt {13} }$$
$$ \Rightarrow \frac {6}{ \sqrt {13} } \hat {j} $$ m/s
magnitude of nel = 2m/s
distance of origin to the line $$ y = \frac {3}{2} \times 12 $$
$$ |d| = | \frac { 0 +0 -2}{ \sqrt { 1+ 9/4} } | \Rightarrow \frac { 2 \times 2 }{ \sqrt {13} } = \frac {y }{ \sqrt {13} } $$
Angular momentum = $$ m \mu r $$
$$ 5 \times 2 \times \frac {4}{ \sqrt {13} } = \frac {40}{ \sqrt {13}} kg \frac {m^2}{s} $$
Two small blocks A of mass $$ \sqrt {3} kg $$ and $$ B $$ of mass kg are connected by a light string are placed on a smooth fixed sphere such that the system is in equilibrium for $$ < AOB = \frac { \pi}{2} $$ then
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The value of $$ \alpha = \frac { \pi}{3} $$ and equilibrium is table
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The value of $$ \alpha = \frac { \pi}{3} $$ and equilibrium is unstable
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The value of $$ \alpha = \frac { \pi}{6} $$ and equilibrium is stable
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The value of $$ \alpha = \frac {\pi}{6} $$ and equilibrium is unstable
Explanation
$$ T = mg sin \alpha $$
$$ T = mg sin ( so \div \alpha) $$
$$ \Rightarrow ma = mg $$
$$ \therefore g sin \alpha = \sqrt {3} g cos \alpha $$
$$ fan \alpha = \sqrt {3} 2 tan \frac { \pi}{3} $$
$$ \therefore \alpha = \frac { \pi}{3} $$ at stable
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