Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Physics-

General

Easy

Question

A pendulum of mass m hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up a plane of inclination with acceleration a0 is

$θ = \tan^{- 1} α$
$θ = \tan^{- 1} (\frac{a_{0}}{g})$
$θ = \tan^{- 1} (\frac{g}{a_{0}})$
$θ = \tan^{- 1} (\frac{a_{0} + g \sin α}{g \cos α})$

The correct answer is: $theta equals tan to the power of negative 1 end exponent open parentheses fraction numerator a subscript 0 plus g sin alpha over denominator g cos alpha end fraction close parentheses$

Related Questions to study

In the situation shown in figure all the string are light and inextensible and pullies are light. There is no friction at any surface and all block are of cuboidal shape. A horizontal force of magnitude F is applied to right most free end of string in both cases of figure 1 and figure 2 as shown. At the instant shown, the tension in all strings are non zero. Let the magnitude of acceleration of large blocks (of mass M) in figure 1 and figure 2 are $a subscript 1$ and $a subscript 2$ respectively. Then:

In the situation shown in figure all the string are light and inextensible and pullies are light. There is no friction at any surface and all block are of cuboidal shape. A horizontal force of magnitude F is applied to right most free end of string in both cases of figure 1 and figure 2 as shown. At the instant shown, the tension in all strings are non zero. Let the magnitude of acceleration of large blocks (of mass M) in figure 1 and figure 2 are $a subscript 1$ and $a subscript 2$ respectively. Then:

physics-General

A plank is held at an angle to the horizontal (Fig.) on two fixed supports A and B. The plank can slide against the supports (without friction) because of its weight Mg. Acceleration and direction in which a man of mass m should move so that the plank does not move.

A plank is held at an angle to the horizontal (Fig.) on two fixed supports A and B. The plank can slide against the supports (without friction) because of its weight Mg. Acceleration and direction in which a man of mass m should move so that the plank does not move.

physics-General

A block of mass m1 lies on top of fixed wedge as shown in figure-1 and another block of mass m2 lies on top of wedge which is free to move as shown in figure-2. At time t = 0, both the blocks are released from rest from a vertical height h above the respective horizontal surface on which the wedge is placed as shown. There is no frcition between block and wedge in both the figures. Let T1 and T2 be the time taken by block in figure-1 and block in figure-2 respectively to just reach the horizontal surface, then :

A block of mass m1 lies on top of fixed wedge as shown in figure-1 and another block of mass m2 lies on top of wedge which is free to move as shown in figure-2. At time t = 0, both the blocks are released from rest from a vertical height h above the respective horizontal surface on which the wedge is placed as shown. There is no frcition between block and wedge in both the figures. Let T1 and T2 be the time taken by block in figure-1 and block in figure-2 respectively to just reach the horizontal surface, then :

physics-General

parallel

System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of cylinder is –

System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of cylinder is –

physics-General

Let $f left parenthesis x right parenthesis equals x c o s to the power of negative 1 end exponent invisible function application left parenthesis negative s i n invisible function application vertical line x vertical line right parenthesis comma x element of open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator pi over denominator 2 end fraction close square brackets$ , then which of the following is true?

Let $f left parenthesis x right parenthesis equals x c o s to the power of negative 1 end exponent invisible function application left parenthesis negative s i n invisible function application vertical line x vertical line right parenthesis comma x element of open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator pi over denominator 2 end fraction close square brackets$ , then which of the following is true?

For all twice differentiable functios $f colon R rightwards arrow R$ , with $f left parenthesis 0 right parenthesis equals f left parenthesis 1 right parenthesis equals f to the power of ´ end exponent left parenthesis 0 right parenthesis equals 0$

For all twice differentiable functios $f colon R rightwards arrow R$ , with $f left parenthesis 0 right parenthesis equals f left parenthesis 1 right parenthesis equals f to the power of ´ end exponent left parenthesis 0 right parenthesis equals 0$

parallel

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $theta$ should be :–

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $theta$ should be :–

physics-General

A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by :–

A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by :–

physics-General

Let $f$ be a twice differentiable function on $left parenthesis 1 , 6 right parenthesis$ . If $f left parenthesis 2 right parenthesis equals 8$ , $f to the power of ´ end exponent left parenthesis 2 right parenthesis equals 5 comma f to the power of ´ end exponent left parenthesis x right parenthesis greater or equal than 1$ and $f to the power of ´ left parenthesis x right parenthesis greater or equal than 4$ , for all $x element of left parenthesis 1 , 6 right parenthesis$ , then $colon$

Let $f$ be a twice differentiable function on $left parenthesis 1 , 6 right parenthesis$ . If $f left parenthesis 2 right parenthesis equals 8$ , $f to the power of ´ end exponent left parenthesis 2 right parenthesis equals 5 comma f to the power of ´ end exponent left parenthesis x right parenthesis greater or equal than 1$ and $f to the power of ´ left parenthesis x right parenthesis greater or equal than 4$ , for all $x element of left parenthesis 1 , 6 right parenthesis$ , then $colon$

parallel

A insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the surface and the insect is $1 third$ . If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of $alpha$ is given :

A insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the surface and the insect is $1 third$ . If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of $alpha$ is given :

physics-General

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with a constant angular acceleration, . If the coefficient of friction between the rod and bead is , and gravity is neglected, then the time after which the bead starts slipping is

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with a constant angular acceleration, . If the coefficient of friction between the rod and bead is , and gravity is neglected, then the time after which the bead starts slipping is

physics-General

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals $mu$ The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction $mu$ and the distance x(=QR), are, respectively close to:

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals $mu$ The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction $mu$ and the distance x(=QR), are, respectively close to:

physics-General

parallel

An observer can see through a pin–hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is

The line of sight of the observer remains constant, making an angle of 45° with the normal.

sin invisible function application theta equals fraction numerator h over denominator square root of h to the power of 2 end exponent plus left parenthesis 2 h right parenthesis to the power of 2 end exponent end root end fraction

fraction numerator 1 over denominator square root of 5 end fraction

mu equals fraction numerator sin invisible function application 4 5 to the power of o end exponent over denominator sin invisible function application theta end fraction

equals fraction numerator 1 divided by square root of 2 over denominator 1 divided by square root of 5 end fraction equals square root of open parentheses fraction numerator 5 over denominator 2 end fraction close parentheses end root

An observer can see through a pin–hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is

physics-General

The line of sight of the observer remains constant, making an angle of 45° with the normal.

sin invisible function application theta equals fraction numerator h over denominator square root of h to the power of 2 end exponent plus left parenthesis 2 h right parenthesis to the power of 2 end exponent end root end fraction

fraction numerator 1 over denominator square root of 5 end fraction

mu equals fraction numerator sin invisible function application 4 5 to the power of o end exponent over denominator sin invisible function application theta end fraction

equals fraction numerator 1 divided by square root of 2 over denominator 1 divided by square root of 5 end fraction equals square root of open parentheses fraction numerator 5 over denominator 2 end fraction close parentheses end root

A diverging beam of light from a point source S having divergence angle a, falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is

Since rays after passing through the glass slab just suffer lateral displacement hence, we have angle between the emergent rays as

alpha

A diverging beam of light from a point source S having divergence angle a, falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is

physics-General

Since rays after passing through the glass slab just suffer lateral displacement hence, we have angle between the emergent rays as

alpha

A rectangular glass slab ABCD, of refractive index n₁, is immersed in water of refractive index $n subscript 2 end subscript open parentheses n subscript 1 end subscript greater than n subscript 2 end subscript close parentheses$ A ray of light in incident at the surface AB of the slab as shown. The maximum value of the angle of incidence a_max, such that the ray comes out only from the other surface CD is given by

A rectangular glass slab ABCD, of refractive index n₁, is immersed in water of refractive index $n subscript 2 end subscript open parentheses n subscript 1 end subscript greater than n subscript 2 end subscript close parentheses$ A ray of light in incident at the surface AB of the slab as shown. The maximum value of the angle of incidence a_max, such that the ray comes out only from the other surface CD is given by

physics-General

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.