Question

Convert newton into dyne by dimensional method​

Answer 1

the dimensional formula of Force F=[M 1 L 1 T −2 ]

the dimensional formula of Force F=[M 1 L 1 T −2 ]So a=1,b=1,c=−2

the dimensional formula of Force F=[M 1 L 1 T −2 ]So a=1,b=1,c=−2Now because Newton is in m.k.s system, we have

the dimensional formula of Force F=[M 1 L 1 T −2 ]So a=1,b=1,c=−2Now because Newton is in m.k.s system, we haveM 1 =1kg

L 1 =1m

L 1 =1mT 1 =1s

L 1 =1mT 1 =1sM 2 =1g

L 2 =1cm

T 2 =1s

T 2 =1sn 1 =1Newton

T 2 =1sn 1 =1Newtonn 2 =?

T 2 =1sn 1 =1Newtonn 2 =?Using the formula : n 2 =n1

n 2 =n1

n 2 =n1 [M1/M2] a

n 2 =n1 [M1/M2] a [L1/L2] b

n 2 =n1 [M1/M2] a [L1/L2] b [T1/T2] c

n 2 =n1 [M1/M2] a [L1/L2] b [T1/T2] c n 2 =1[1kg/1g] 1

n 2 =n1 [M1/M2] a [L1/L2] b [T1/T2] c n 2 =1[1kg/1g] 1 [1m/1cm] 1 [1s/1s] −2

n 2 =n1 [M1/M2] a [L1/L2] b [T1/T2] c n 2 =1[1kg/1g] 1 [1m/1cm] 1 [1s/1s] −2 n 2 =1[1000g/1g] 1 [100cm/1cm] =1×1

1×1n 2 =1000×100=100000=10

5 Dynes.

Hi, hope it may be helpful

Answer 2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Lets say that 1N = x Dynes.

Now we know that Force is

${m_1}^{1} {l_1}^{1} {t_1}^{ - 2}$

And dyne is

${m_2}^{1} {l_2}^{1} {t_2}^{ - 2}$

Further conversion is in the attachment.

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BY SCIVIBHANSHU

THANK YOU

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