Advertisements
Advertisements
प्रश्न
Make L the subject of formula T = `2pisqrt("L"/"G")`
Advertisements
उत्तर
T = `2pisqrt("L"/"G")`
⇒ `"T"/(2pi) = sqrt("L"/"G"`
squaring both sides
⇒ `("T"/(2pi))^2 = "L"/"G"`
⇒ `"G"("T"/(2pi))^2` = L
⇒ L = `("GT"^2)/(4pi^2)`.
APPEARS IN
संबंधित प्रश्न
Make N the subject of formula I = `"NG"/("R" + "Ny")`
Make d the subject of formula S = `"n"/(2){2"a" + ("n" - 1)"d"}`
Make R2 the subject of formula R2 = 4π(R12 - R22)
Make c the subject of formula x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"a")`
Make h the subject of the formula K = `sqrt("hg"/"d"^2 - "a"^2`. Find h, when k = -2, a = -3, d = 8 and g = 32.
Make f the subject of the formula D = `sqrt((("f" + "p")/("f" - "p"))`. Find f, when D = 13 and P = 21.
"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.
"The volume of a cone V is equal to the product of one third of π and square of radius r of the base and the height h". Express this statement as a formula. Make r the subject formula. Find r, when V = 1232cm3, π = `(22)/(7)`, h = 24cm.
If s = `"n"/(2)[2"a" + ("n" - 1)"d"]`, the n express d in terms of s, a and n. find d if n = 3, a = n + 1 and s = 18.
"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.
