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Physics-

General

Easy

Question

Portion AB of the wedge shown in figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. The ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point C is :

3/4
5
7/5
8/3

The correct answer is: 5

Related Questions to study

Inner and outer radii of a spool are r and R respectively. A thread is wound over its inner surface and placed over a rough horizontal surface. Thread is pulled by a force F as shown in fig. then in case of pure rolling

Inner and outer radii of a spool are r and R respectively. A thread is wound over its inner surface and placed over a rough horizontal surface. Thread is pulled by a force F as shown in fig. then in case of pure rolling

physics-General

Two identical circular loops are moving with same kinetic energy one rolls & other slides. The ratio of their speed is :

Two identical circular loops are moving with same kinetic energy one rolls & other slides. The ratio of their speed is :

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A disc of radius R is rolling purely on a flat horizontal surface, with a constant angular velocity. The angle between the velocity and acceleration vectors of point P is

A disc of radius R is rolling purely on a flat horizontal surface, with a constant angular velocity. The angle between the velocity and acceleration vectors of point P is

physics-General

parallel

A thin rod of length L is placed at angle $theta$ to vertical on a frictionless horizontal floor and released. If the center of mass has acceleration = A, and the rod an angular acceleration = $alpha$ at initial moment, then

A thin rod of length L is placed at angle $theta$ to vertical on a frictionless horizontal floor and released. If the center of mass has acceleration = A, and the rod an angular acceleration = $alpha$ at initial moment, then

physics-General

If $f$ is a real-valued differentiable function satisfying $vertical line f left parenthesis x right parenthesis minus f left parenthesis y right parenthesis vertical line less or equal than left parenthesis x minus y right parenthesis to the power of 2 end exponent comma x comma y element of R$ and f(0)=0, then f(1) equals

If $f$ is a real-valued differentiable function satisfying $vertical line f left parenthesis x right parenthesis minus f left parenthesis y right parenthesis vertical line less or equal than left parenthesis x minus y right parenthesis to the power of 2 end exponent comma x comma y element of R$ and f(0)=0, then f(1) equals

If f(x) = $open curly brackets table row cell fraction numerator x to the power of 2 end exponent minus 1 over denominator x to the power of 2 end exponent plus 1 end fraction end cell cell comma 0 less than x less or equal than 2 end cell row cell fraction numerator 1 over denominator 4 end fraction open parentheses x to the power of 3 end exponent minus x to the power of 2 end exponent close parentheses end cell cell comma 2 less than x less or equal than 3 end cell row cell fraction numerator 9 over denominator 4 end fraction left parenthesis vertical line x minus 4 vertical line plus vertical line 2 minus x vertical line right parenthesis comma end cell cell 3 less than x less than 4 end cell end table comma close$ then:

If f(x) = $open curly brackets table row cell fraction numerator x to the power of 2 end exponent minus 1 over denominator x to the power of 2 end exponent plus 1 end fraction end cell cell comma 0 less than x less or equal than 2 end cell row cell fraction numerator 1 over denominator 4 end fraction open parentheses x to the power of 3 end exponent minus x to the power of 2 end exponent close parentheses end cell cell comma 2 less than x less or equal than 3 end cell row cell fraction numerator 9 over denominator 4 end fraction left parenthesis vertical line x minus 4 vertical line plus vertical line 2 minus x vertical line right parenthesis comma end cell cell 3 less than x less than 4 end cell end table comma close$ then:

parallel

where represents the integral part function, then:

where represents the integral part function, then:

Which of the following function(s) has/have removable discontinuity at

Which of the following function(s) has/have removable discontinuity at

Consider the function f(x)= $open square brackets table row cell x left curly bracket x right curly bracket plus 1 end cell cell 0 less or equal than x less than 1 end cell row cell 2 minus left curly bracket x right curly bracket end cell cell 1 less or equal than x less or equal than 2 end cell end table close$ where {x} denotes the fractional part function. Which one of the following statements is NOT correct?

Consider the function f(x)= $open square brackets table row cell x left curly bracket x right curly bracket plus 1 end cell cell 0 less or equal than x less than 1 end cell row cell 2 minus left curly bracket x right curly bracket end cell cell 1 less or equal than x less or equal than 2 end cell end table close$ where {x} denotes the fractional part function. Which one of the following statements is NOT correct?

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Statement-I : If $a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma horizontal ellipsis horizontal ellipsis a subscript n end subscript greater than 0$ , then $l i m subscript x rightwards arrow infinity end subscript open curly brackets fraction numerator a subscript 1 end subscript superscript 1 divided by x end superscript plus a subscript 2 end subscript superscript 1 divided by x end superscript plus a subscript 3 end subscript superscript 1 divided by x end superscript plus horizontal ellipsis.. plus a subscript n end subscript superscript 1 divided by x end superscript over denominator n end fraction close curly brackets to the power of n x end exponent equals product subscript i equals 1 end subscript superscript n end superscript a subscript n end subscript$

Statement-I : If $a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma horizontal ellipsis horizontal ellipsis a subscript n end subscript greater than 0$ , then $l i m subscript x rightwards arrow infinity end subscript open curly brackets fraction numerator a subscript 1 end subscript superscript 1 divided by x end superscript plus a subscript 2 end subscript superscript 1 divided by x end superscript plus a subscript 3 end subscript superscript 1 divided by x end superscript plus horizontal ellipsis.. plus a subscript n end subscript superscript 1 divided by x end superscript over denominator n end fraction close curly brackets to the power of n x end exponent equals product subscript i equals 1 end subscript superscript n end superscript a subscript n end subscript$

$L i m subscript x rightwards arrow 0 end subscript invisible function application fraction numerator left parenthesis 1 minus c o s invisible function application 2 x right parenthesis left parenthesis 3 plus c o s invisible function application x right parenthesis over denominator x t a n invisible function application 4 x end fraction$ is equal to

$L i m subscript x rightwards arrow 0 end subscript invisible function application fraction numerator left parenthesis 1 minus c o s invisible function application 2 x right parenthesis left parenthesis 3 plus c o s invisible function application x right parenthesis over denominator x t a n invisible function application 4 x end fraction$ is equal to

$l i m subscript s rightwards arrow pi divided by 2 end subscript fraction numerator left square bracket 1 minus t a n invisible function application x divided by 2 right square bracket left square bracket 1 minus s i n invisible function application x right square bracket over denominator left square bracket 1 plus t a n invisible function application x divided by 2 right square bracket left square bracket pi minus 2 x right square bracket to the power of 3 end exponent end fraction$ is equal to-

$l i m subscript s rightwards arrow pi divided by 2 end subscript fraction numerator left square bracket 1 minus t a n invisible function application x divided by 2 right square bracket left square bracket 1 minus s i n invisible function application x right square bracket over denominator left square bracket 1 plus t a n invisible function application x divided by 2 right square bracket left square bracket pi minus 2 x right square bracket to the power of 3 end exponent end fraction$ is equal to-

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Statement $negative I colon$ The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400) is 8000.
Statement - II : $sum subscript k equals 1 end subscript superscript n end superscript open parentheses k to the power of 3 end exponent minus left parenthesis k minus 1 right parenthesis to the power of 3 end exponent close parentheses equals n to the power of 3 end exponent$ , for any natural number n.

Statement $negative I colon$ The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400) is 8000.
Statement - II : $sum subscript k equals 1 end subscript superscript n end superscript open parentheses k to the power of 3 end exponent minus left parenthesis k minus 1 right parenthesis to the power of 3 end exponent close parentheses equals n to the power of 3 end exponent$ , for any natural number n.

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately.

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately.

The negation of the statement "If I become a teacher, then I will open a school", is :

The negation of the statement "If I become a teacher, then I will open a school", is :

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