Question

a stretched string produces a transverse wave of length 0.2m, its frequency of vibration is 1700hz. find the velocity of wave.​

Answer 1

Answer

Science Solutions for Class 9 Cbse class 9 Chapter 1 Motion are provided here with simple step-by-step explanations. These solutions for Motion are extremely popular among Class 9 students for Cbse class 9 Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science Book of Class 9 Cbse class 9 Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Science Solutions. All Science Solutions for class Class 9 Cbse class 9 are prepared by experts and are 100% accurate.

Answer 2

Answer:

AnswEr :

Let the Number be 100.

Increase% = 30%

Decrease% 30%

• New Number will be :

\begin{lgathered}:\implies\sf New \:Number = Old \:Number \times (100 + Increase)\% \times (100 - Decrease)\% \\\\\\:\implies\sf New\:Number = 100 \times (100 + 30)\% \times (100 - 30)\%\\\\\\:\implies\sf New\:Number = 100 \times 130\% \times 70\%\\\\\\:\implies\sf New\:Number = 100 \times\dfrac{130}{100} \times \dfrac{70}{100}\\\\\\:\implies\sf New\:Number = 13 \times 7\\\\\\:\implies\sf New\:Number = 91\end{lgathered}

:⟹NewNumber=OldNumber×(100+Increase)%×(100−Decrease)%

:⟹NewNumber=100×(100+30)%×(100−30)%

:⟹NewNumber=100×130%×70%

:⟹NewNumber=100×

100

130

×

100

70

:⟹NewNumber=13×7

:⟹NewNumber=91

• Percentage Change in the Number :

\begin{lgathered}\longrightarrow\tt Change\% = \dfrac{Change}{Old\:Number}\times100\\\\\\\longrightarrow\tt Change\% = \dfrac{(New-Old) Number}{Old\:Number}\times100\\\\\\\longrightarrow\tt Change\% = \dfrac{(100-91)}{100}\times100\\\\\\\longrightarrow \boxed{ \red{\tt Change\% = 9\% \:Decrease}}\end{lgathered}

⟶Change%=

OldNumber

Change

×100

⟶Change%=

OldNumber

(New−Old)Number

×100

⟶Change%=

100

(100−91)

×100

⟶

Change%=9%Decrease

⠀

∴ There will 9% Decrease in the Number.