Consider the frequency distribution of the given number

If the mean is known to be ' 3 , then the value of f is

Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m

Given below is a graph between a variable force (F) (along y-axis) and the displacement (X) (along $x minus$ axis) of a particle in one axis) of a particle in one dimension. The work done by the force in the displacement interval between 0 m and 30 m is

Velocity time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all forces on the particle is

A toy car of mass 4 kg moves up a ramp under the influence of force $F$ plotted against displacement x. The maximum height attained is given (Take $open g equals 10 text end text m divided by s to the power of 2 end exponent close parentheses$

If Wa, Wb, and Wc represent the work done in moving a particle from X to Y along three different path a, b, and c respectively (as shown) in the gravitational field of a point mass m, find the correct relation between Wa, Wb and Wc

Let $P equals 3 to the power of 1 divided by 3 end exponent times 9 to the power of 1 divided by 9 end exponent times 27 to the power of 1 divided by 27 end exponent horizontal ellipsis horizontal ellipsis infinity$ then $P to the power of 1 divided by 3 end exponent equals$

Let $a subscript 1 end subscript comma a subscript 2 end subscript horizontal ellipsis.. a subscript 10 end subscript$ be A.P and $h subscript 1 end subscript comma h subscript 2 end subscript horizontal ellipsis h subscript 10 end subscript$ be in H.P If $a subscript 1 end subscript equals h subscript 1 end subscript equals 2$ and $a subscript 10 end subscript equals h subscript 10 end subscript equals 3$ the $a subscript 3 end subscript h subscript 8 end subscript equals$

The work done by a force acting on a body is as shown in the graph. what is the total work done in covering an initial distance of 15 m?

A 8 kg mass moves along x - axis. Its accelerations as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x=0 to x=6 cm ?

If A and G are A.M and G.M between two positive numbers a and b are connected by the relation A+G=a-b then the numbers are in the ratio

We have to be careful while solving quadratic equation questions as there are chances of mistakes with the signs while finding the roots. Sometimes the students try to use factorisation methods to solve the quadratic equation, but in this type of questions, it is not recommended at all. Always try to use the quadratic formula for solving. You can also remember the statement to be proved, given in the question as a property for two positive numbers.

If A and G are A.M and G.M between two positive numbers a and b are connected by the relation A+G=a-b then the numbers are in the ratio

If $x equals sum subscript n equals 0 end subscript superscript infinity end superscript a to the power of n end exponent comma y equals sum subscript n equals 0 end subscript superscript infinity end superscript b to the power of n end exponent comma z equals sum subscript n equals 0 end subscript superscript infinity end superscript left parenthesis a b right parenthesis to the power of n end exponent$ where a<1, b<1 then

If l, m, n are the $p to the power of text th end text end exponent comma q to the power of text th end text end exponent comma r to the power of text th end text end exponent$ terms of a G.P which are +ve, then $open vertical bar table attributes columnalign left end attributes row cell l o g invisible function application l end cell p 1 row cell l o g invisible function application m end cell q 1 row cell l o g invisible function application n end cell r 1 end table close vertical bar equals$

The relationship between force and position is shown in the figure given (in one dimensional case) calculate the work done by the force in displacing a body from x=0 cm to x=5 cm

Fifth term of G.P is 2 The product of its first nine terms is

Here note that the fifth term is having fourth power of 2 and not fifth power. We need not to find all nine terms separately; only finding the product is enough because that product will then be written in the form of the term that is known. Terms in a G.P. are having a common ratio in between. That’s why the power of r is increasing as the terms are increasing.

Fifth term of G.P is 2 The product of its first nine terms is