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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.
1. |
Two bodies (M) and (N) of equal masses are suspended from two separate massless springs of spring constants (k_1) and (k_2) respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of (M) to the of (N) is.A. `(k_(1))/(k_(2))`B. `sqrt((k_(1))/(k_(2)))`C. `(k_(2))/(k_(1))`D. `sqrt((k_(2))/(k_(1)))` |
Answer» Correct Answer - D<br>Maximum velocity`=a omega = asqrt((k)/(m))`<br> Given that `omega = sqrt((k_(1))/(m)) = a_(2)sqrt((k_(1))/(m))rArr (a_(1))/(a_(2)) = sqrt((k_(2))/(k_(1)))` | |
2. |
Two identacal rods each of length `l` and mass `m` weided toeather at right angle and edge suspended from a kinetic sides as shown Angular frequency of small oscillation of the system in its then plane about the total of suspension is A. `sqrt((3g)/(4sqrt(2)l))`B. `sqrt((3g)/(2sqrt(2)l))`C. `sqrt((3g)/(sqrt(2)l))`D. None of these |
Answer» Correct Answer - B<br>`C` is combined center of mass<br> <br> ` h = (1)/(2) cos 45^(@) = (1)/(2sqrt(2))`<br> moments of ineetia of rods about `O : 1 = (2ml^(2))/(3)`<br> `T = 2pi sqrt((1)/(2mgh))`<br> ` 2pi = sqrt(2ml^(2)sqrt(2))/(3[2mg]l) = 2pi sqrt((2sqrt(2l))/(3g))`<br> Angular frequancy omega `= (2pi)/(T) = sqrt((3g)/(2sqrt(2l)))` | |
3. |
The period of oscillation of a simple pendulum of length (L) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination (prop), is given by.A. `2pi sqrt((L)/(g cos alpha))`B. `2pi sqrt((L)/(g sin alpha))`C. `2pi sqrt((L)/(g))`D. `2pi sqrt((L)/(g tan alpha))` |
Answer» Correct Answer - A<br>See the following force diagram<br> <br>Vehicle is moving down the frictionless inclined surfacwe so, its acceleration is `g sin theta` Since vehicle is acceleration apseudo force `m(g sintheta)` will act on bob of pendulum which cancels the `sintheta ` component of weight of the bob.Hence net force on the bob is `F_(net) = mg cos theta ` or net acceleration of the bob is `g_(eff) = g cos theta`<br> `:.` Time period `T = 2 pi sqrt((1)/(g_(eff))) = 2pi sqrt((1)/( g cos theta))` | |
4. |
Two very small having mass `m` are atteched to two masses rods of length `l` Now these rods are joined it form `V` like figure having angle `60^(@)` This assumility is new higest in a verticle plane so that it can rotan without any friction about a horizontal axis perpendicular in the plane of figure) as shown in the figure The period of small oscilation of this asseamble is A. `2pi sqrt((2l)/(g))`B. `2pi sqrt((2l)/(sqrt(3)g))`C. `2pi sqrt((1)/(sqrt(3)g))`D. `2pisqrt((sqrt(3)l)/(g))` |
Answer» Correct Answer - B<br>`T = 2pi sqrt((2ml^(2))/(2mgl))cos 30^(@) = 2pi sqrt((2l)/(gsqrt(3))` | |
5. |
A uniform square plate at side `a` is hinged at one at its comes it is suspended such than it can rotate about horizontal axis. The time period of small oscillation about its equilibrium position.A. `2pi sqrt((2a)/(3g))`B. `2 pi sqrt((sqrt(2)a)/(3g))`C. `2pisqrt((2sqrt(2)a)/(3g))`D. `2 pi sqrt((2sqrt(2a))/(g))` |
Answer» Correct Answer - C<br><br> `T = 2pi sqrt((1)/(mgd)) = 2pi sqrt((m((a^(2) + a^(2)))/(12) + m((a)/(sqrt(2)))^(2))/(mg"(a)/(sqrt(2))))`<br> `= 2pi sqrt(((a)/(6) + (a)/(2)) .sqrt(2))/(g) = 2pi sqrt(2sqrt(2)a)/(3g)` | |
6. |
A body of mass `m` is atteched to the lower end of a spring whose upper end is fixed .The spring has negaligible mass .When the mass `m` is slightly puylled down and released it oscillation with a time period of `3 s` when the mass `m` is increased by `1kg` time period of oscillations becomes `5s` The value of `m` in `kg` isA. `(16)/(9)`B. `(9)/(16)`C. `(3)/(4)`D. `(4)/(3)` |
Answer» Correct Answer - B<br>`T = 2pi sqrt((m)/(k))`<br> `3 = 2 pi sqrt((m)/(k))` ….(i)<br> `5 = 2 pi sqrt((m+1)/(k))`….(ii)<br> `((1)^(2))/((2)^(2)) rArr (9)/(19) = (m)/(m + 1) rArr m = (9)/(16)` | |
7. |
What will be the period of the displacement body of mass `m`? A. `2pisqrt((m)/(2K))`B. `2pisqrt((3m)/(K))`C. `2pisqrt((3m)/(2K))`D. `pisqrt((3m)/(K))` |
Answer» Correct Answer - C<br>Equivalent force constant of three spring<br> `k_(eq) = (k_(1)k_(2))/(k_(1) + k_(2)) = (k xx 2k)/(3k) = (2k)/(3)`<br> `T = 2pi sqrt((m)/(k_(eq))) = 2pi sqrt((m)/(2k//3))`<br> `rArr T = 2pi sqrt((3m)/(2k))` | |
8. |
Time period of a block when suspended from the upper plate of a parallel plate capactor by a spring of stiffness `k` is when block ids unchanged If a change `g` is given to the block then new time period of oscilation will be A. `T`B. `gtT`C. `ltT`D. `ge T` |
Answer» Correct Answer - A<br>The force that act on the block are `qE mg `and spring force since `qE` and `mg` are constant force the only variable elactic force change by `kx` where `x` is elongation in the spring<br> Unbalanced (restoring) force<br> `F = - kx`<br> `or - m omega^(2) x = - kx`<br> `or, omega = sqrt((k)/(m))`<br> `T = (2m)/(omega) = 2pi sqrt((m)/(k))` | |