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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 36951. |
Select odd man out(incorrect) from the following expressions for kinetic energy in pure rolling. |
| Answer» <html><body><p>In pure <a href="https://interviewquestions.tuteehub.com/tag/rolling-1190754" style="font-weight:bold;" target="_blank" title="Click to know more about ROLLING">ROLLING</a> K.E. `=(K.E.)_("<a href="https://interviewquestions.tuteehub.com/tag/trans-1425477" style="font-weight:bold;" target="_blank" title="Click to know more about TRANS">TRANS</a>")+(K.E.)_("Rot")`<br/>In pure rolling K.E. `=1/2 mv_(CM)^(2)+1/2I_(CM)omega^(2)`<br/>In pure rolling `K.E. =1/2mv^(2)` <br/>In pure rolling kinetic energy with reference to centre of mass is K.E. `=1/2mv_(CM)^(2)(1+K^(2)//<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>^(2))`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :It is an incorrect expression. It is not for pure rolling.</body></html> | |
| 36952. |
Answer the following questions : (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provided angular magnification ? (b) In viewing through a magnifying glass, one usually positions one's eye very close to the lens. Does angular magnification change if the eye is moved back ? ( c) Magnifying power of a simple a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power ? (d) Why must both the objective and eye piece of a compound microscope have a short focal lengths ? (e) when viewing through a compound microscope, our eyes should be positioned not on the eye piece, but a short distance away from it for best veiwing, why ? How much should be that short distance betweenthe eye and eye piece ? |
| Answer» <html><body><p></p>Solution :(a) It is true that angular size of imahe is equal to angular size of the object. <br/> By using magnifying glass, we keep the object far more closer to the eye than at `25 cm`, its normal position without use of glass. The closer object has <a href="https://interviewquestions.tuteehub.com/tag/larger-1067345" style="font-weight:bold;" target="_blank" title="Click to know more about LARGER">LARGER</a> angular size than the same object at `25 cm`. It is in this sense that angular magnification is achieved. <br/> (b) Yes, the angular magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less than the angle subtended at the lens. The <a href="https://interviewquestions.tuteehub.com/tag/effect-966056" style="font-weight:bold;" target="_blank" title="Click to know more about EFFECT">EFFECT</a> is negligible when image is at much larger distance. <br/> ( c) Theoretically, it is true. However, when we decrease focal length, aberrations both spherical and chromatic become more pronounced. Further, it is difficult to grind lenses of very small focal lengths. <br/> (d) Angular magnification of eye piece is `(1 + (d)/(f_e))`. This increases as `f_e` decreases. <br/> Further, magnification of objective lens is `(v)/(u)`. As object <a href="https://interviewquestions.tuteehub.com/tag/lies-1073086" style="font-weight:bold;" target="_blank" title="Click to know more about LIES">LIES</a> close to focus of objective lens `u ~~ f_0`. To increase this magnification `(v//f_0), f_0` should be smaller. <br/> (e) The image of objective lens in eye piece is called 'eye ring'. All the rays from the object <a href="https://interviewquestions.tuteehub.com/tag/refracted-7709131" style="font-weight:bold;" target="_blank" title="Click to know more about REFRACTED">REFRACTED</a> by the objective go through the eye ring. Therefore, ideal position for our eyes for viewing is this eye ring only. When eye is too close to the eye piece, <a href="https://interviewquestions.tuteehub.com/tag/field-987291" style="font-weight:bold;" target="_blank" title="Click to know more about FIELD">FIELD</a> of view reduces and eyes fo not collect much of the light. The precise location of the eye would depend upon the separation between the obeject and eye piece, and also on focal length of the eye piece.</body></html> | |
| 36953. |
A block of wood of mass 0.5kg is placed on plane making angle 30° with the horizontal. If the coefficient of friction between the surface of contact of the body and the plane is 0.2, what force is required to keep the body sliding down with uniform velocity? |
| Answer» <html><body><p>`6.4N`<br/>`1.6N`<br/>`3.2N`<br/>`4.8N` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36954. |
An artificial satellite of the earth releases a packet. If air resistance is neglected, the point where the packet will hit, will be |
| Answer» <html><body><p>a head<br/>exactly below<br/>behind<br/>it will <a href="https://interviewquestions.tuteehub.com/tag/never-570518" style="font-weight:bold;" target="_blank" title="Click to know more about NEVER">NEVER</a> <a href="https://interviewquestions.tuteehub.com/tag/reach-1178062" style="font-weight:bold;" target="_blank" title="Click to know more about REACH">REACH</a> the earth</p>Answer :D</body></html> | |
| 36955. |
A planet revolves round the sun in an elliptical orbit of minor and major axes x and y respectively. Then the time period of revolution is proportional to |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>+y)^(3//2)`<br/>`(y-x)^(3//2)`<br/>`x^(3//2)`<br/>`y^(3//2)` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 36956. |
Two identical point masses are fixed on Y-axis at equal distances from the origin. A small particle starts on the X-axis from x = oo towards the origin. Its acceleration a is taken positive along its direction of motion. Out of the following graphs the one that best represents the variation of a with x-coordinate a particle is |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_DOC_OBJ_PHY_XI_V01_B_C09_E01_095_O01.png" width="30%"/><br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_DOC_OBJ_PHY_XI_V01_B_C09_E01_095_O02.png" width="30%"/><br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_DOC_OBJ_PHY_XI_V01_B_C09_E01_095_O03.png" width="30%"/><br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_DOC_OBJ_PHY_XI_V01_B_C09_E01_095_O04.png" width="30%"/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 36957. |
A person standing in the centre of a room, 12m high looks into a plane mirror fixed on the wall. Then the minimum length of the plane mirror for him to see the full length image of the wall behing him, is equal to |
| Answer» <html><body><p>12m<br/>8m<br/>4m<br/>6m</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36958. |
The amplitude of vibration of a particle is given by a_(m)=(a_(0))/(aw^(2)-bw+c) where a_(0), a , b and c are positive. The condition for a simple resonant frequency |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=<a href="https://interviewquestions.tuteehub.com/tag/4ac-318871" style="font-weight:bold;" target="_blank" title="Click to know more about 4AC">4AC</a>`<br/>`b^(2)gt4ac`<br/>`b^(2)=5ac`<br/>`b^(2)<a href="https://interviewquestions.tuteehub.com/tag/lt-537906" style="font-weight:bold;" target="_blank" title="Click to know more about LT">LT</a> 4ac`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 36959. |
A particle is moving with shm in a straight line. When the distance of the particle from the equilibrium position has the value x_1 and x_2the corresponding values of velocities are v_1and v_2show that period is T=2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2) |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> at a <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> `x_1` is `v_1=omegasqrt(a^2-x_1^2)` <br/> and `v_2=<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a> <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(a^2-x_2^2)` <br/><a href="https://interviewquestions.tuteehub.com/tag/squaring-3058178" style="font-weight:bold;" target="_blank" title="Click to know more about SQUARING">SQUARING</a> and subtracting <br/> `v_1^2-v_2^2=omega^2[a^2-x_1^2-a^2+x_2^2]` <br/> `omega^2=(v_1^2-v_2^2)/(x_2^2-x_1^2)` <br/>`omega=[(v_1^2-v_2^2)/(x_2^2-x_1^2)]^(1//2)` <br/> `(2pi)/T=omega"or " T=(2pi)/omega` <br/>`T = 2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)`</body></html> | |
| 36960. |
A and B are two satellites revolving round the earth in circular orbits have time periods 8hr and 1 hr respectively . The ratio of their radius of orbits |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>^(3//2) : <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>`<br/>`8:1`<br/>`4:1`<br/>`2:1`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36961. |
A block of mass m is pulled along a horizontal surface by applying a force at an angle theta with the horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is mu, then the work done by the applied force is |
| Answer» <html><body><p>`(mu mgd)/(cos theta + mu sin theta)` <br/>`(mu m g d cos theta)/(cos theta + mu cos theta)` <br/>`(mu m g d sin theta)/(cos theta + mu sin theta)`<br/>`(mu m g d cos theta)/(cos theta - mu sin theta)` </p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C04_E01_001_S01.png" width="80%"/> <br/> Because the block moves with a uniform velocity, the resultant force is zero. Resolving <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> into horizontal components `F cos theta` and vertical component `F sin theta,` we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> <br/> `R + F sin theta = mg orR = mg - F sin theta` <br/> Also , `f = mu R = mu(mg - F sin theta)` <br/> But `f = F cos theta` <br/> `:. F cos theta = mu (mg - F sin theta)` or `F (cos theta + mu sin theta) = mumg` <br/> or `F = (mu mg)/(cos theta + mu sin theta theta)` <br/> work done, `<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> = <a href="https://interviewquestions.tuteehub.com/tag/fs-1000993" style="font-weight:bold;" target="_blank" title="Click to know more about FS">FS</a> cos theta, W= (mu m g d cos theta)/(cos theta + mu sin theta) ( :. s = d)`</body></html> | |
| 36962. |
How are we able to break a wire by repeated bending ? |
| Answer» <html><body><p>It is an easy process<br/>Repeated <a href="https://interviewquestions.tuteehub.com/tag/bending-895097" style="font-weight:bold;" target="_blank" title="Click to know more about BENDING">BENDING</a> cause <a href="https://interviewquestions.tuteehub.com/tag/elastic-967540" style="font-weight:bold;" target="_blank" title="Click to know more about ELASTIC">ELASTIC</a> fatigue<br/>It is a <a href="https://interviewquestions.tuteehub.com/tag/wrong-1462035" style="font-weight:bold;" target="_blank" title="Click to know more about WRONG">WRONG</a> question.<br/>None</p>Solution :Repeated bending of wire decreases elastic strength and <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> it can be broken <a href="https://interviewquestions.tuteehub.com/tag/easily-964537" style="font-weight:bold;" target="_blank" title="Click to know more about EASILY">EASILY</a>.</body></html> | |
| 36963. |
What is 'Force constant' of a spring? On what factors does it depend? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Force constant (k) of a spring is defined as the ratio of force <a href="https://interviewquestions.tuteehub.com/tag/applied-380664" style="font-weight:bold;" target="_blank" title="Click to know more about APPLIED">APPLIED</a> to displacement <a href="https://interviewquestions.tuteehub.com/tag/produced-592947" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCED">PRODUCED</a> in the spring. Force constant `k= F/y`. Force constant k depends on the elastic properties of spring <a href="https://interviewquestions.tuteehub.com/tag/material-1089170" style="font-weight:bold;" target="_blank" title="Click to know more about MATERIAL">MATERIAL</a>.</body></html> | |
| 36964. |
A constant force F is applied at the top of a ring of that M and radius is R as shown. The angular momentum of particle about point constact at time |
| Answer» <html><body><p>is <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/increases-1040626" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASES">INCREASES</a> <a href="https://interviewquestions.tuteehub.com/tag/linearly-2157962" style="font-weight:bold;" target="_blank" title="Click to know more about LINEARLY">LINEARLY</a> with time<br/>is 2FRt<br/>decreases linearly with time </p>Answer :B::C</body></html> | |
| 36965. |
A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel force of lever arm l. One force is of magnitude F and acts at one extreme end. The length of the rod is (x(F+"ma")l)/("ma"). Where 'x' is |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a></body></html> | |
| 36966. |
Arrange the following simple pendulum in ascending order of their periods of oscillation a.of length 1 m at a place of g= 8m//s b.of length 2m at a place of g=10m//s c. of length 4m at a place of g= 4m//s |
| Answer» <html><body><p>a, <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>. <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a><br/>b, a , c <br/>b, c, a<br/>c, b, a</p>Answer :A</body></html> | |
| 36967. |
A mass attached to a spring is free to oscillate, with angular velocity omega, in a horizontal plane without friction or damping. It is pulled to a distance x_0 and pushed towards the centre with a velocity v_0 at time t=0. Determine the amplitude of the resulting oscillaters omega, x_(0)" and "v_(0). [Hint : Start with the equation x= a cos (omega t+theta) andnote that the initial velocity is negative.] |
| Answer» <html><body><p></p>Solution :Displacement at time t in SHM, <br/> `x = a cos (omega t+phi)"""…….."(1)` <br/> Where `phi` = initial phase, `omega` = angular <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> and a= amplitude <br/> <a href="https://interviewquestions.tuteehub.com/tag/differentiating-953151" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENTIATING">DIFFERENTIATING</a> equation (1) w.r.t. .t., <br/> `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>= (dx)/(dt)` <br/> `=(d)/(dt) [a cos (omega t+phi)]` <br/> `therefore v = -a omega sin (omega t+ phi)""".........."(2)` <br/> When `t=0, x =x_(0)" and "(dx)/(dt)= -v_(0)` <br/> From equation (1), <br/> `x_(0) = a cos (omega xx 0+phi)` <br/> `x_(0) = a cos phi ""........."(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> and from equation (2), <br/> `-v_(0) = a omega sin (0+phi)""[t=0]` <br/> `therefore v_(0) = a omega sin phi` <br/> `therefore (v_0)/(omega) = a sin phi"""........."(4)` <br/> Squaring and <a href="https://interviewquestions.tuteehub.com/tag/adding-2399902" style="font-weight:bold;" target="_blank" title="Click to know more about ADDING">ADDING</a> equation (3) and (4), <br/> `x_(0)^(2)+(v_(0)^(2))/(omega^2)= a^(2)cos^(2) phi+ a^(2) sin^(2) phi` <br/> `=a^(2) (cos^(2) phi + sin^(2) phi)` <br/> `therefore x_(0)^(2) +(v_(0)^(2))/(omega^2) = a^(2)""[therefore cos^(2) phi+ sin^(2) phi= 1]` <br/> `therefore a= sqrt(x_(0)^(2) +(v_(0)^(2))/(omega^2))`.</body></html> | |
| 36968. |
Assertion: The difference of two vectors A and B can be treated as the sum of two vectors. Subtraction of vectors can be defined in terms of addition of vectors. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Let two <a href="https://interviewquestions.tuteehub.com/tag/vectors-14090" style="font-weight:bold;" target="_blank" title="Click to know more about VECTORS">VECTORS</a> be `vecA and <a href="https://interviewquestions.tuteehub.com/tag/vecb-3257249" style="font-weight:bold;" target="_blank" title="Click to know more about VECB">VECB</a>`. <a href="https://interviewquestions.tuteehub.com/tag/difference-951394" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENCE">DIFFERENCE</a> of vectors `vecA and vecB` can be takenas the sum of two vectors `vecA and -vecB`. i.e., `vecA - vecB = vecA = (-vecB)`</body></html> | |
| 36969. |
Three liquids A, B and C having same specific heat and mass , 2m and 3m have temperatures 20^(@)C, 40^(@)C and 60^(@)C respectively. Temperature of the mixture when (MPP_PHY_C13_E01_274_Q01.png" width="80%"> |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :(A) T, (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>)<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>, (<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) <a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>, (D) T</body></html> | |
| 36970. |
Whichof the following pairs is a correctpairA particle revolves round a circular path the acceleration of the particle is |
| Answer» <html><body><p>along the circumference of the circle and `a=(<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(m)`<br/>alongthe tangent and v= <a href="https://interviewquestions.tuteehub.com/tag/ar-380980" style="font-weight:bold;" target="_blank" title="Click to know more about AR">AR</a> <br/>along the radius and `a=(v^(2))/(r )`<br/>zero and <a href="https://interviewquestions.tuteehub.com/tag/velocityis-3259015" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITYIS">VELOCITYIS</a> zero </p>Solution :Accelerationactstowardsthe centre(where= v =velocitym- <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> r-radiusofcircularpatha -acceleration)</body></html> | |
| 36971. |
Prove the law of conservation of linear momentum use it to find the recoil velocity ofa gun when a bullet is fired from it |
| Answer» <html><body><p></p>Solution :(i) The forceon eachparticle ( Newton'ssecondlaw) can be written as <br/> ` vecF_(<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>) = ( dvecp_(1))/(dt)and vecF_(21) = ( dvecP_(2))/(dt)`<br/>(ii)Here ` vecP_(1)` is the <a href="https://interviewquestions.tuteehub.com/tag/momentumof-2840779" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUMOF">MOMENTUMOF</a> particle1 whichchanges due to the <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a>` vecF_(12)`exertedby particle 2. Further ` vecP_(2)`is the momentumof particle 2. <br/>Thischanges due to ` vecF_(21)`exertedby particle 1 . <br/>` ( dvecp_(1))/(dt) = - (dvecp_(2))/( dt)` <br/>` ( dvecP_(1))/( dt)+ ( d vecP_(2))/(dt) = 0` <br/>` d/( dt)( vecP_(1) + vecP_(2)) = 0`<br/> (iii) It impliesthat ` vecP_(1) + vecP_(2)` = constantvector ( always) <br/> (iv) `vecP_(1) + vecP_(2) ` is the totallinearmomentumof thetwoparticles` ( vecP_(" tot"= vecP_(1) + vecP_(2))` . It is also called as total linear momentumof the <a href="https://interviewquestions.tuteehub.com/tag/system-1237255" style="font-weight:bold;" target="_blank" title="Click to know more about SYSTEM">SYSTEM</a>. Here , the two particles constitutethe system. <br/> (v) Ifthere are no externalforcesactingon thesystem,then the totallinear momentumof thesystem ` (vecP_(" tot"))`is alwaysa constantvector .</body></html> | |
| 36972. |
A body executing S.H.M has maximum acceleration equal to 24m//"sec" and maximum velocity equal to 16m//"sec". The amplitude of S.H.M is |
| Answer» <html><body><p>`(3)/(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)m`<br/>`(64)/(<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>)m`<br/>`(32)/(3)m`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/1024-266309" style="font-weight:bold;" target="_blank" title="Click to know more about 1024">1024</a>)/(9)m`</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36973. |
A particle falls towards the earth from infinity. The velocity with which it reaches earth's surface (g to acceleration due to gravity on earth's surface and R is the radius of the earth) |
| Answer» <html><body><p>`2gR`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(2gR)`<br/>`sqrt(<a href="https://interviewquestions.tuteehub.com/tag/gr-1010452" style="font-weight:bold;" target="_blank" title="Click to know more about GR">GR</a>)`<br/>`R"/"<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>`</p>Answer :B</body></html> | |
| 36974. |
A ball of mass m moving with a speed v makes a head on collision with an identical ball at rest. The kinetic energy after collision of the balls is three fourth of the original kinetic energy. The coefficient of restitution is |
| Answer» <html><body><p>`1/2`<br/>`1/3`<br/>`1/(sqrt2)`<br/>`1/(sqrt3)`</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C04_E01_127_S01.png" width="80%"/> <br/> Applying the principle of conservation of linear momentum , we get <br/>`mv = mv_1 + mv_2 " or " v = v_1 + v_2 "" ….(i)` <br/>By <a href="https://interviewquestions.tuteehub.com/tag/defination-947003" style="font-weight:bold;" target="_blank" title="Click to know more about DEFINATION">DEFINATION</a> of coefficient of restitution <br/>`e = (v_2 - v_1)/(u_1 - u_2) = (v_2 - v_1)/(v - 0)" or " v_2 - v_1 = ev "" .....(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>)` <br/> As KE after <a href="https://interviewquestions.tuteehub.com/tag/collision-922060" style="font-weight:bold;" target="_blank" title="Click to know more about COLLISION">COLLISION</a> = `3/4 KE` before collision <br/>`:. 1/2 m (v_1^2 + v_2^2) = 3/4 xx 1/2 mv^2 " or " v_1^2 + v_2^2 = 3/4 v^2 "".....(iii)` <br/> Squaring eq. (i), we get `v_1^2 + v_2^2 + 2v_1 v_2 = v^2 "".........(<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>)` <br/>Subtracting (iiii), from <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (iv), we get <br/>`2v_1 v_2 = 1/4 v^2"".....(v)` <br/> Squaring (ii), we get <br/> `v_2^2 + v_1^2 - 2v_1 v_2 = e^2 v^2 "" ....(vi)` <br/>Using (iii) and (v) in (vi), we get <br/> `3/4 v^2 - 1/4 v^2 = e^2 v^2 , 1/2 v^2 = e^2 v^2 or e = 1/(sqrt2)`.</body></html> | |
| 36975. |
A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plots shown, indicae the one the represents the velocity (v) of the pebble as a function of time (t). |
| Answer» <html><body><p> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_PMH_C11_E01_092_O01.png" width="30%"/><br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_PMH_C11_E01_092_O02.png" width="30%"/><br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_PMH_C11_E01_092_O03.png" width="30%"/><br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_PMH_C11_E01_092_O04.png" width="30%"/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36976. |
Select the correct alternative: In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :total liner <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> and total <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of the system of two <a href="https://interviewquestions.tuteehub.com/tag/bodies-900171" style="font-weight:bold;" target="_blank" title="Click to know more about BODIES">BODIES</a> .</body></html> | |
| 36977. |
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh^(-1).Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^(-1). What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] |
| Answer» <html><body><p></p>Solution :Average velocity `= ("Total displacement")/("Time interval")` <br/> and average speed `= ("Path length")/("Time interval")` <br/> (i) For 0 to 30 minutes time interval, <br/> Distance covered from home to <a href="https://interviewquestions.tuteehub.com/tag/market-25464" style="font-weight:bold;" target="_blank" title="Click to know more about MARKET">MARKET</a> , `=2.5 km` <br/> Speed =5 km/hr <br/> `therefore` Time taken in market `= (2.5)/(5) = 0.5hours = 30` minutes <br/> (a) Average velocity `= ("displacement")/("time") = (2.5)/(0.5) =6km//hr` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) Average speed `= (2.5)/(0.5) = 5 km//hr` <br/> (ii) For 0 to 50 minutes time interval, <br/> Velcity while returning back `=7.5 km//hr` <br/> distance `=2.5 km` <br/> `therefore ` Time taken to cover this distance `= (2.5)/(7.5) = 1/3` <a href="https://interviewquestions.tuteehub.com/tag/hours-1030090" style="font-weight:bold;" target="_blank" title="Click to know more about HOURS">HOURS</a> = <a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> minutes <br/> Time taken to complete the travelling `=30+20 ` (home to market and market to home) <br/> = 50 minutes`=5/6` hours <br/> Displacement `=2.5 -2.5 =0 km` <br/> `therefore` Average velocity `= (0)/(5//6)=0` <br/> Total distance covered `=2.5 + 2.5 =5 km` <br/> Average speed `= (5)/(5//6) =6` km/hr <br/> (iii) For time interval 0 to 40 minutes, <br/> Time taken by man to reach market from home, <br/> =30 minutes <br/> `therefore ` Time to return back, <br/> `= 40 -30= 10` minutes <br/> Distance covered in 10 minuutes, <br/> = velocity while returning `xx` time <br/> `=7.5 xx (10)/(60) = 1.25 km` <br/> Displacement `=2.5 - 1.25 =1.25 km` <br/> `therefore ` Average velocity `= (1.25)/(40//60)=1.875km//hr` <br/> Avereage speed `= ("total distance")/("time")` <br/> `=(2.5 +1.25)/(40//60) =3.75 xx (60)/(40)=5.625` km/hr <br/> In second case, average velocity is zero but not average speed.</body></html> | |
| 36978. |
Three spherical balls of masses 1kg, 2kg and 3kg are placed at the corners of an equilateral triangle of side 1m. The magnitude of the gravitational force exterted by 2 kg and 3kg masses on 1 kg mass. |
| Answer» <html><body><p>19 G<br/>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>)G`<br/>`sqrt(19)G`<br/>`(G)/(sqrt(17))`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36979. |
A body of mass 5kg is thrown up vertically with a kinetic energy of 1000J. If acceleration due to gravity is 10ms^(-2), find the height at which the kinetic energy becomes half of the original value. |
| Answer» <html><body><p>10m <br/><a href="https://interviewquestions.tuteehub.com/tag/20m-293046" style="font-weight:bold;" target="_blank" title="Click to know more about 20M">20M</a> <br/>50 m <br/>100 m </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> : Half KE is converted into <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> energy . <br/> then `PE =1/2 KE` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/mgh-1095458" style="font-weight:bold;" target="_blank" title="Click to know more about MGH">MGH</a> = 1000/2 = 500` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = (500)/(mg) = (500)/(5 xx 10) = 10m `</body></html> | |
| 36980. |
Find the maximum speed at which a car can turn round a curve of 36 m radius on a level road. Given the coefficient of friction between the tyre and the road is 0.53. |
| Answer» <html><body><p></p>Solution :Radius of the <a href="https://interviewquestions.tuteehub.com/tag/curve-941741" style="font-weight:bold;" target="_blank" title="Click to know more about CURVE">CURVE</a> r= 36 m<br/>Coefficient of <a href="https://interviewquestions.tuteehub.com/tag/friction-21545" style="font-weight:bold;" target="_blank" title="Click to know more about FRICTION">FRICTION</a> `mu`= 0.53<br/>Acceleration due to gravity `g= 10 ms^(-2)` <br/> `v_(<a href="https://interviewquestions.tuteehub.com/tag/max-546895" style="font-weight:bold;" target="_blank" title="Click to know more about MAX">MAX</a>) = sqrt(mu <a href="https://interviewquestions.tuteehub.com/tag/rg-1188560" style="font-weight:bold;" target="_blank" title="Click to know more about RG">RG</a>) = sqrt( 0.53 xx 36 xx 10 ) = 13.81 ms^(-1)`</body></html> | |
| 36981. |
A force F is given by F at + bt^(2) where t is time. What are dimensions of 'a' and 'b'? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/mlt-2163963" style="font-weight:bold;" target="_blank" title="Click to know more about MLT">MLT</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` and `ML^2T^4` <br/>`MLT^(-3)` and `MLT^(-4)` <br/>`MLT^(-1) ` and `MLT^0` <br/>`MLT^(-4) ` and `MLT^1`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36982. |
It is a well known fact that during a toal solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather fro examples 2.3and 2.4, determine the approximate diameter of the moon. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :3,581 <a href="https://interviewquestions.tuteehub.com/tag/km-1064498" style="font-weight:bold;" target="_blank" title="Click to know more about KM">KM</a></body></html> | |
| 36983. |
A motor is rotating at a constant angular velocity of 600 rpm. The angular displacement in 2 second is ………….. |
| Answer» <html><body><p>`40pi` rad<br/>`20pi` rad<br/>`<a href="https://interviewquestions.tuteehub.com/tag/10pi-1774733" style="font-weight:bold;" target="_blank" title="Click to know more about 10PI">10PI</a>` rad<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>)/(3)<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>` rad</p>Solution :`omega=600" rpm"=(600xx2pi)/(60)=20pi" rad/s"` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> A=omegat=20pixx2=40pi" rad"`</body></html> | |
| 36984. |
K is the force constant of a spring. The work done is increasing its extension from l_(1), to l_(2), will be |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>(l_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)-l_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))`<br/>`(K)/(2)(l_(2)+l_(1))`<br/>`K(l_(2)^(2)-l_(1)^(2))`<br/>`(K)/(2)(l_(2)^(2)-l_(1)^(2))`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 36985. |
Define frame of reference and give its types. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/frame-998974" style="font-weight:bold;" target="_blank" title="Click to know more about FRAME">FRAME</a> of reference : The place or <a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a> from which <a href="https://interviewquestions.tuteehub.com/tag/observer-1127626" style="font-weight:bold;" target="_blank" title="Click to know more about OBSERVER">OBSERVER</a> observes the situation (motion) is called from of reference. It has <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> <a href="https://interviewquestions.tuteehub.com/tag/types-15918" style="font-weight:bold;" target="_blank" title="Click to know more about TYPES">TYPES</a>. <br/> (1) Inertial (frame to Reference <br/>(2) Non-inertial (frame to) Reference</body></html> | |
| 36986. |
The gravitational force is always attractive in nature. Why? |
| Answer» <html><body><p></p>Solution :Because of negative sign in equation `<a href="https://interviewquestions.tuteehub.com/tag/vec-723433" style="font-weight:bold;" target="_blank" title="Click to know more about VEC">VEC</a>(F) = -(GM_(1)M_(2))/(r^2) hat^(r)` that the gravitational force is always attractive in nature and the direction of the force is along the <a href="https://interviewquestions.tuteehub.com/tag/line-1074199" style="font-weight:bold;" target="_blank" title="Click to know more about LINE">LINE</a> <a href="https://interviewquestions.tuteehub.com/tag/joining-525969" style="font-weight:bold;" target="_blank" title="Click to know more about JOINING">JOINING</a> the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> masses.</body></html> | |
| 36987. |
Which of the following statements are ture/ false ? (i) Keplar discovered famous theory of relativity. (ii) Nuclear reactors are based on controlled nuclear chain reaction. (iii) Einstein explained photoemission on the basis of Planck's quantum theory. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(i) <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) True</body></html> | |
| 36988. |
A satellite is moving in a circular orbit around the earth with a speed equal to half the escape speed from the earth. If 'R' is the radius of the earth, then the height of the satellite above the surface of the earth is |
| Answer» <html><body><p>R/2<br/>2R/3<br/>R<br/>2R</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36989. |
The lower end of a capillary tube ofdiameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker . What is the pressure required in the tube in order to blow a hemispherical bubble at the end in water ?The surface tension of water at temperature of the experiments is 7.30xx10^(-2)Nm^(9-1). 1 atmospheric pressure =1.01xx10^(5)Pa.Density of water =1000kg//m^(3),g=9.80ms^(-2).Also calulate the excess pressure . |
| Answer» <html><body><p></p>Solution :Radius of capillary tube ,<br/>`r=(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.0)/(2)mm`<br/>`=1.0mm=<a href="https://interviewquestions.tuteehub.com/tag/1xx10-1804645" style="font-weight:bold;" target="_blank" title="Click to know more about 1XX10">1XX10</a>^(-3)m`<br/>Surface tension of water `=7.30xx10^(-2)Nm^(-1)`<br/>Atmospheric <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> `P=1.01xx10^(5)Pa`<br/>Density of water `rho=1000kgm^(-3)`<br/>Acceleration of gravity `g=9.8ms^(-2)`<br/><a href="https://interviewquestions.tuteehub.com/tag/depth-948801" style="font-weight:bold;" target="_blank" title="Click to know more about DEPTH">DEPTH</a> of bubble `h=8.0cm=8xx10^(-2)m`<br/>The pressure difference between inside and outside the bubble form in liquid, <br/> `P_(i)-P_(o)=(2s)/(r)`<br/>The radius if bubble is equal to the radius of capillary tube)<br/>`thereforeP_(i)=P_(o)+(2s)/(r)` ...(1)<br/>But `P_(o)=P_(a)+hrhog`<br/>`=1.01xx10^(5)+8xx10^(-2)xx10^(3)xx9.8`<br/>`=1.01xx10^(5)+784`<br/>`1.01784xx10^(5)Nm^(-2)`<br/>From equation (1),<br/>`P_(i)=P_(o)+(2s)/(r)`<br/>`=1.01784xx10^(5)+(2xx7.3xx10^(-2))/(1xx10^(-3))`<br/>`=1.01784xx10^(5)+14.6xx10`<br/>`=1.01784+146`<br/>`=1.01930`<br/>`=1.02xx10^(5)N//m^(2)`<br/>The excess pressure in a bubble , `P_(i)-P_(o)=1.01930xx10^(5)-1.01784xx10^(5)`<br/>`=0.00146xx10^(5)`<br/>`=146Pa`</body></html> | |
| 36990. |
The temperature of 5 moles of a gas at constantvolume is changed from 100^(@)C to 120^(@)C. The change in internal energy is 80J. The total heat capacity of the gas at constant volume will be in joule/kelvin is |
| Answer» <html><body><p>8<br/>4<br/>0.8<br/>0.4</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36991. |
A particle executes SHM has amplitude 6cm. Its acceleration at a distance from the mean position 2cm is 8cm//sec^(2). The maximum speed of the particle is |
| Answer» <html><body><p>`8cm//"<a href="https://interviewquestions.tuteehub.com/tag/sec-1197209" style="font-weight:bold;" target="_blank" title="Click to know more about SEC">SEC</a>"`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/12cm-271618" style="font-weight:bold;" target="_blank" title="Click to know more about 12CM">12CM</a>//"sec"`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/16cm-277845" style="font-weight:bold;" target="_blank" title="Click to know more about 16CM">16CM</a>//"sec"`<br/>`24cm//"sec"`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :B</body></html> | |
| 36992. |
Moment of inertia of a circular ring of mass m and radius r about the normal axis passing through its centre is |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/mr-549185" style="font-weight:bold;" target="_blank" title="Click to know more about MR">MR</a>^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(4)`<br/>`mr^(2)`<br/>`(mr^(2))/(2)`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)/(2) mr^(2)`</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36993. |
For vectors vec(A) = (2i + 3j - 2k), vec(B) = (5i + nj + k) and vec(C ) = (-I + 2j + 3k) to be coplanar, the value of n is |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a></body></html> | |
| 36994. |
A man of 60 kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what height he can jump by using this energy? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/4m-318788" style="font-weight:bold;" target="_blank" title="Click to know more about 4M">4M</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/20m-293046" style="font-weight:bold;" target="_blank" title="Click to know more about 20M">20M</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/28m-299762" style="font-weight:bold;" target="_blank" title="Click to know more about 28M">28M</a> <br/>0.2 m</p>Answer :A</body></html> | |
| 36995. |
The maximum restoring force of a body executing SHM is alpha and total energy is beta obtain it’s the amplitude in terms of beta" and "alpha. |
| Answer» <html><body><p></p>Solution :Maximum restoring force `F= <a href="https://interviewquestions.tuteehub.com/tag/ka-528931" style="font-weight:bold;" target="_blank" title="Click to know more about KA">KA</a>` <br/> `therefore alpha = kA` <br/> Total energy `<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>= (1)/(2) kA^(2)` <br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/beta-2557" style="font-weight:bold;" target="_blank" title="Click to know more about BETA">BETA</a> = (1)/(2) kA^(2)` <br/> `therefore (beta)/(alpha)= (A)/(2)` <br/> `therefore A= (2beta)/(alpha)`.</body></html> | |
| 36996. |
A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released in starts performing simple harmonic motion at angular frequency omega. If the radius of the bottle is 2.5 cm then omega is close |
| Answer» <html><body><p>`2.50 rad"/"s`<br/>`3.75 rad"/"s`<br/>`5.00 rad"/"s`<br/>`1.25 rad"/"s`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to law of floatation, <br/> (Weight of <a href="https://interviewquestions.tuteehub.com/tag/floating-992903" style="font-weight:bold;" target="_blank" title="Click to know more about FLOATING">FLOATING</a> object)= (Weight of <a href="https://interviewquestions.tuteehub.com/tag/liquid-1075124" style="font-weight:bold;" target="_blank" title="Click to know more about LIQUID">LIQUID</a> displaced by floating object) <br/> `therefore mg = m_(0)g` <br/> `therefore m= m_(0) implies V. p_(0)= Alp_(0)` <br/> `implies l= (m)/(A p_(0))= (m)/(pi R^(2) p_(0))` <br/> `therefore l= (310g)/((3.14)(2.5 cm)^(2)(1-(g)/(cm^3))) approx <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> cm approx 0.16 m` <br/> Now angular frequency, <br/> `omega = sqrt((g)/(l)) = sqrt((9.8)/(0.16))= sqrt(61.25)= 7.826 (rad)/(s)` <br/> The nearest value in <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> options for above answer is `5 rad"/"s`. <br/> `therefore omega approx 5.0 (rad)/(s)`. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P2_C14_E05_066_S01.png" width="80%"/></body></html> | |
| 36997. |
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :3675 Non each <a href="https://interviewquestions.tuteehub.com/tag/front-1000768" style="font-weight:bold;" target="_blank" title="Click to know more about FRONT">FRONT</a> wheel, 5145 N on each <a href="https://interviewquestions.tuteehub.com/tag/back-389278" style="font-weight:bold;" target="_blank" title="Click to know more about BACK">BACK</a> wheel.</body></html> | |
| 36998. |
A closed cubical box is completely filled wilth water and is accelerated horizontally towards right with an acceleration a. The resultant normal force by the water on the top of the box. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/passes-1148517" style="font-weight:bold;" target="_blank" title="Click to know more about PASSES">PASSES</a> through the centre of the top<br/>Passes through a point to the right of the centre<br/>Passes through a point to the left of the centre<br/>Becomes zero</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36999. |
Describe the measurement of Earth's shadow (umbra) radius during total lunar eclipse. |
| Answer» <html><body><p></p>Solution :(i) It is possible to measure the radius of shawdow of the Earth at the point where the Moon crosses. <br/> (ii) When the Moon is inside the umbra shadow, it appears red in <a href="https://interviewquestions.tuteehub.com/tag/colour-422259" style="font-weight:bold;" target="_blank" title="Click to know more about COLOUR">COLOUR</a>. As soon as the Moon exits from the umbra shadow, it appears in crescent shape. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) By finding the apparent radii of the Earth's Umbra shadow and the Moon, the <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> of the these radii can be <a href="https://interviewquestions.tuteehub.com/tag/calculated-907694" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATED">CALCULATED</a>. This is shown in figure.<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_PHY_XI_V02_C06_E03_016_S01.png" width="80%"/> <br/> (iv) Schematic diagram of umbra disk radius. The apparent radius of Earth's umbra shadow = `R_(S) = 13.2 cm` <br/> The apparent radius of the Moon = `R_(m) = 5.15 cm` <br/> (v) The ratio `(R_s)/(R_m) = 2.56` <br/> The radius of the Earth's umbra shadow is <br/> `R_(s) = 2.56 xx R_(m)`.</body></html> | |
| 37000. |
A particle of mass.m. is projected with a velocity .u. at an angle . theta. with the horizontal. Work done by gravity during its descent from its highest point to a point which is at half the maximum height is |
| Answer» <html><body><p>0<br/>`(mu^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a>^(2)theta)/4`<br/>`1/2mu^(2) sin^(2)theta`<br/>`1/2mu^(2)cos^(2)theta` </p>Answer :B</body></html> | |