Advertisements
Advertisements
प्रश्न
Make k the subject of formula T = `2pisqrt(("k"^2 + "h"^2)/"hg"`
उत्तर
T = `2pisqrt(("k"^2 + "h"^2)/"hg"`
⇒ `"T"/(2pi) = sqrt(("k"^2 + "h"^2)/"hg"`
Squaring both sides
⇒ `("T"/(2pi))^2 = ("k"^2 + "h"^2)/"hg"`
⇒ `"hg"("T"/(2pi))^2 - "h"^2` = k2
⇒ k = `sqrt(("T"^2"hg")/(4pi^2) - "h"^2`.
APPEARS IN
संबंधित प्रश्न
Make x the subject of formula `"a"x^2/"a"^2 + y^2/"b"^2` = 1
Make r2 the subject of formula `(1)/"R" = (1)/"r"_1 + (1)/"r"_2`
Make N the subject of formula I = `"NG"/("R" + "Ny")`
Given: mx + ny = p and y = ax + b. Find x in terms of m, n, p, a and b.
If A = pr2 and C = 2pr, then express r in terms of A and C.
Make a the subject of the formula S = `"n"/(2){2"a" + ("n" - 1)"d"}`. Find a when S = 50, n = 10 and d = 2.
Make y the subject of the formula `x/"a" + y/"b" `= 1. Find y, when a = 2, b = 8 and x = 5.
Make g the subject of the formula v2 = u2 - 2gh. Find g, when v = 9.8, u = 41.5 and h = 25.4.
"The volume of a cylinder V is equal to the product of π and square of radius r and the height h". Express this statement as a formula. Make r the subject formula. Find r, when V = 44cm3, π = `(22)/(7)`, h = 14cm.
"Area A oof a circular ring formed by 2 concentric circles is equal to the product of pie and the difference of the square of the bigger radius R and the square of the bigger radius R and the square of the smaller radius r. Express the above statement as a formula. Make r the subject of the formula and find r, when A = 88 sq cm and R = 8cm.
