Position of a body with acceleration a is given by $x = k a^{m} t^{n}$ , here t is time, find the dimension of m and n.

m = 1, n = 1

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m = 1, n = 2

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m = 2, n = 1

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m = 2, n = 2

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Solution

The correct option is C $m = 1, n = 2$
Position $x = K a^{m} t^{n}$
Writing the dimension on the both sides.
$[M^{0} L T^{0}] = [L T^{- 2}]^{m} [T]^{n}$
$[M^{0} L^{m} T^{- 2 m + n}]$
On comparing both sides.
$m = 1$
$- 2 m + n = 0$
$n = 2 m = 2 \times 1 = 2$ .

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If the ratio of sum of m terms and n terms of an A.P. is , m²:n² then prove that the ratio of its mth and nth terms will be 2m-1 : 2n-1.

I know the answer for this, but i want to confirm if this method is right because everywhere, the other method is given!

S(m)÷s(n) = m²/n²
On solving till a point, we get this equation

2an + mnd -nd = 2am + mnd -dm
→ 2an-nd = 2am-dm
→ n(2a-d) = m(2a-d)
→ n=m. ......(1)

In the 'to prove' statement,

LHS →
a+(m-1)d / a+(n-1)d

Replacing n with m (using 1)

a+(m-1)d/a+(m-1)d = 1/1 = 1

RHS →

2m-1/2n-1
Replacing n with m (using 1)
2m-1/2m-1
= 1/1
=1

Hence, LHS=RHS.

IS THIS CORRECT?

The coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) externally in the ratio m : n are given by

Q. If ^mC₁= ⁿC₂ , then
(a) 2 m = n
(b) 2 m = n (n + 1)
(c) 2 m = n (n − 1)
(d) 2 n = m (m − 1)

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Position of a body with acceleration a is given by x=kamtn, here t is time, find the dimension of m and n.

The correct option is C m=1,n=2Position x=KamtnWriting the dimension on the both sides.[M0LT0]=[LT−2]m[T]n[M0LmT−2m+n]On comparing both sides.m=1−2m+n=0n=2m=2×1=2.

Position of a body with acceleration a is given by $x = k a^{m} t^{n}$ , here t is time, find the dimension of m and n.

The correct option is C $m = 1, n = 2$
Position $x = K a^{m} t^{n}$
Writing the dimension on the both sides.
$[M^{0} L T^{0}] = [L T^{- 2}]^{m} [T]^{n}$
$[M^{0} L^{m} T^{- 2 m + n}]$
On comparing both sides.
$m = 1$
$- 2 m + n = 0$
$n = 2 m = 2 \times 1 = 2$ .