Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

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 stone of mass $1 blank k g$ tied to a light inextensible string of length $L equals fraction numerator 10 over denominator 3 end fraction m$ is whirling in a circular path of radius $L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if $g$ is taken to be $10 m divided by sec to the power of 2 end exponent invisible function application blank comma$ the speed of the stone at the highest point of the circle is

$20 m / s e c$
$10 \sqrt{3} m / s e c$
$5 \sqrt{2} m / s e c$
$10 m / s e c$

The correct answer is: $10 m divided by s e c$

Since the maximum tension $T subscript B end subscript$ in the string moving in the vertical circle is at the bottom and minimum tension $T subscript T end subscript$ is at the top
$therefore T subscript B end subscript equals fraction numerator m v subscript B end subscript superscript 2 end superscript over denominator L end fraction plus m g$ and $T subscript T end subscript equals fraction numerator m v subscript T end subscript superscript 2 end superscript over denominator L end fraction minus m g$
$therefore fraction numerator T subscript B end subscript over denominator T subscript T end subscript end fraction equals fraction numerator fraction numerator m v subscript B end subscript superscript 2 end superscript over denominator L end fraction plus m g over denominator fraction numerator m v subscript T end subscript superscript 2 end superscript over denominator L end fraction minus m g end fraction equals fraction numerator 4 over denominator 1 end fraction o r blank fraction numerator v subscript B end subscript superscript 2 end superscript plus g L over denominator v subscript T end subscript superscript 2 end superscript minus g L end fraction equals fraction numerator 4 over denominator 1 end fraction$
Or $v subscript B end subscript superscript 2 end superscript plus g L equals 4 v subscript T end subscript superscript 2 end superscript minus 4 g L$ but $v subscript B end subscript superscript 2 end superscript equals v subscript T end subscript superscript 2 end superscript plus 4 g L$
$therefore v subscript T end subscript superscript 2 end superscript plus 4 g L plus g L equals 4 v subscript T end subscript superscript 2 end superscript minus 4 g L rightwards double arrow 3 v subscript T end subscript superscript 2 end superscript equals 9 g L$
$therefore v subscript T end subscript superscript 2 end superscript equals 3 cross times g cross times L equals 3 cross times 10 cross times fraction numerator 10 over denominator 3 end fraction o r v subscript T end subscript equals 10 blank m divided by s e c$

Related Questions to study

In a circus stuntman rides a motorbike in a circular track of radius $R$ in the vertical plane. The minimum speed at highest point of track will be

In a circus stuntman rides a motorbike in a circular track of radius $R$ in the vertical plane. The minimum speed at highest point of track will be

Physics-General

If a cycle wheel of radius 4 $m$ completes one revolution in two seconds. Then acceleration of a point on the cycle wheel will be

If a cycle wheel of radius 4 $m$ completes one revolution in two seconds. Then acceleration of a point on the cycle wheel will be

Physics-General

A cart is moving horizontally along a straight line with constant speed $30 blank m divided by s$ . A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved $80 blank m$ . At what speed (relative to the cart) must the projectile be fired $left parenthesis$ Take $blank g equals 10 blank m divided by s to the power of 2 end exponent right parenthesis$

A cart is moving horizontally along a straight line with constant speed $30 blank m divided by s$ . A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved $80 blank m$ . At what speed (relative to the cart) must the projectile be fired $left parenthesis$ Take $blank g equals 10 blank m divided by s to the power of 2 end exponent right parenthesis$

Physics-General

parallel

A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a subscript c end subscript$ is varying with time $t$ as, $a subscript c end subscript equals k to the power of 2 end exponent r t to the power of 2 end exponent$ , The power delivered to the particle by the forces acting on it is

A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a subscript c end subscript$ is varying with time $t$ as, $a subscript c end subscript equals k to the power of 2 end exponent r t to the power of 2 end exponent$ , The power delivered to the particle by the forces acting on it is

Physics-General

A particle is moving in a circle of radius $R$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t equals 0 blank i s blank v subscript 0 end subscript$ , the time taken to complete the first revolution is

A particle is moving in a circle of radius $R$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t equals 0 blank i s blank v subscript 0 end subscript$ , the time taken to complete the first revolution is

Physics-General

A particle of mass $m$ is rotating in a horizontal circle of radius $R$ and is attached to a hanging mass $M$ as shown in the figure. The speed of rotation required by the mass $m$ keep $M$ steady is

A particle of mass $m$ is rotating in a horizontal circle of radius $R$ and is attached to a hanging mass $M$ as shown in the figure. The speed of rotation required by the mass $m$ keep $M$ steady is
Physics-General

parallel

A constant force $stack F with rightwards arrow on top$ = 3 $stack i with hat on top$ + 2 $stack j with hat on top$ – 4 $stack k with hat on top$ is applied at point A (1, –1, 2). Then vector moment of $stack F with rightwards arrow on top$ about point B (2,–1, 3) is -

A constant force $stack F with rightwards arrow on top$ = 3 $stack i with hat on top$ + 2 $stack j with hat on top$ – 4 $stack k with hat on top$ is applied at point A (1, –1, 2). Then vector moment of $stack F with rightwards arrow on top$ about point B (2,–1, 3) is -

The vectors $stack i with hat on top$ + 2 $stack j with hat on top$ +3 $stack k with hat on top$ ,  $stack i with hat on top$ + 4 $stack j with hat on top$ + 7 $stack k with hat on top$ &–3 $stack i with hat on top$ –2 $stack j with hat on top$ –5 $stack k with hat on top$ are collinear, if  =

The vectors $stack i with hat on top$ + 2 $stack j with hat on top$ +3 $stack k with hat on top$ ,  $stack i with hat on top$ + 4 $stack j with hat on top$ + 7 $stack k with hat on top$ &–3 $stack i with hat on top$ –2 $stack j with hat on top$ –5 $stack k with hat on top$ are collinear, if  =

If $stack u with rightwards arrow on top$ , $stack v with rightwards arrow on top$ , $stack w with rightwards arrow on top$ are non-coplanar vectors then
( $stack u with rightwards arrow on top$ + $stack v with rightwards arrow on top$ – $stack w with rightwards arrow on top$ ). ( $stack u with rightwards arrow on top$ – $stack v with rightwards arrow on top$ ) × ( $stack v with rightwards arrow on top$ – $stack w with rightwards arrow on top$ )=

If $stack u with rightwards arrow on top$ , $stack v with rightwards arrow on top$ , $stack w with rightwards arrow on top$ are non-coplanar vectors then
( $stack u with rightwards arrow on top$ + $stack v with rightwards arrow on top$ – $stack w with rightwards arrow on top$ ). ( $stack u with rightwards arrow on top$ – $stack v with rightwards arrow on top$ ) × ( $stack v with rightwards arrow on top$ – $stack w with rightwards arrow on top$ )=

parallel

If a,b,c are different non- negative numbers and a $stack i with hat on top$ + a $stack j with hat on top$ + c $stack k with hat on top$ ; $stack i with hat on top$ + $stack k with hat on top$ and c $stack i with hat on top$ + c $stack j with hat on top$ + b $stack k with hat on top$ lie in a plane, then c is -

If a,b,c are different non- negative numbers and a $stack i with hat on top$ + a $stack j with hat on top$ + c $stack k with hat on top$ ; $stack i with hat on top$ + $stack k with hat on top$ and c $stack i with hat on top$ + c $stack j with hat on top$ + b $stack k with hat on top$ lie in a plane, then c is -

Value of ( $stack a with rightwards arrow on top$ – $stack b with rightwards arrow on top$ ).[( $stack b with rightwards arrow on top$ – $stack c with rightwards arrow on top$ ) × ( $stack c with rightwards arrow on top$ – $stack a with rightwards arrow on top$ )] =

Value of ( $stack a with rightwards arrow on top$ – $stack b with rightwards arrow on top$ ).[( $stack b with rightwards arrow on top$ – $stack c with rightwards arrow on top$ ) × ( $stack c with rightwards arrow on top$ – $stack a with rightwards arrow on top$ )] =

If $stack u with rightwards arrow on top$ , $stack v with rightwards arrow on top$ , $stack w with rightwards arrow on top$ are vectors, such that $stack u with rightwards arrow on top$ + $stack v with rightwards arrow on top$ + $stack w with rightwards arrow on top$ = 0 where | $stack u with rightwards arrow on top$ | = 3, | $stack v with rightwards arrow on top$ | = 4 & | $stack w with rightwards arrow on top$ | = 5 then $stack u with rightwards arrow on top$ . $stack v with rightwards arrow on top$ + $stack v with rightwards arrow on top$ . $stack w with rightwards arrow on top$ + $stack w with rightwards arrow on top$ . $stack u with rightwards arrow on top$ =

If $stack u with rightwards arrow on top$ , $stack v with rightwards arrow on top$ , $stack w with rightwards arrow on top$ are vectors, such that $stack u with rightwards arrow on top$ + $stack v with rightwards arrow on top$ + $stack w with rightwards arrow on top$ = 0 where | $stack u with rightwards arrow on top$ | = 3, | $stack v with rightwards arrow on top$ | = 4 & | $stack w with rightwards arrow on top$ | = 5 then $stack u with rightwards arrow on top$ . $stack v with rightwards arrow on top$ + $stack v with rightwards arrow on top$ . $stack w with rightwards arrow on top$ + $stack w with rightwards arrow on top$ . $stack u with rightwards arrow on top$ =

parallel

If $stack a with rightwards arrow on top$ × $stack b with rightwards arrow on top$ = $stack c with rightwards arrow on top$ × $stack b with rightwards arrow on top$  $stack 0 with rightwards arrow on top$ then -

If $stack a with rightwards arrow on top$ × $stack b with rightwards arrow on top$ = $stack c with rightwards arrow on top$ × $stack b with rightwards arrow on top$  $stack 0 with rightwards arrow on top$ then -

If $stack a with rightwards arrow on top$ and $stack b with rightwards arrow on top$ are two unit vectors such that ( $stack a with rightwards arrow on top$ + 2 $stack b with rightwards arrow on top$ ) & (5 $stack a with rightwards arrow on top$ – 4 $stack b with rightwards arrow on top$ ) are perpendicular to each other then angle between $stack a with rightwards arrow on top$ and $stack b with rightwards arrow on top$ is-

If $stack a with rightwards arrow on top$ and $stack b with rightwards arrow on top$ are two unit vectors such that ( $stack a with rightwards arrow on top$ + 2 $stack b with rightwards arrow on top$ ) & (5 $stack a with rightwards arrow on top$ – 4 $stack b with rightwards arrow on top$ ) are perpendicular to each other then angle between $stack a with rightwards arrow on top$ and $stack b with rightwards arrow on top$ is-

If $stack a with rightwards arrow on top$ = $stack i with hat on top$ + $stack j with hat on top$ + $stack k with hat on top$ & $stack a with rightwards arrow on top$ . $stack b with rightwards arrow on top$ = 1 and $stack a with rightwards arrow on top$ × $stack b with rightwards arrow on top$ = $stack j with hat on top$ – $stack k with hat on top$ . Then $stack b with rightwards arrow on top$ =

If $stack a with rightwards arrow on top$ = $stack i with hat on top$ + $stack j with hat on top$ + $stack k with hat on top$ & $stack a with rightwards arrow on top$ . $stack b with rightwards arrow on top$ = 1 and $stack a with rightwards arrow on top$ × $stack b with rightwards arrow on top$ = $stack j with hat on top$ – $stack k with hat on top$ . Then $stack b with rightwards arrow on top$ =

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.