Math, asked by nayanarawal27, 10 months ago

4. एक लम्बे बाँस का - भाग लाल, भाग हरा, भाग
पीला रंगा है और शेष भाग सफेद है. यदि सफेद भाग
की लम्बाई 4 मीटर हो, तो पूरे बाँस की लम्बाई ज्ञात
करो-
(A) 8
(B) 10
(C) 9
(D) 12​

Answers

Answered by mitajoshi11051976
0

Step-by-step explanation:

{ \huge \bf { \mid{ \overline{ \underline{Correct \: Question}}} \mid}}∣

CorrectQuestion

A bullet of mass 20 grams is horizontally fired with a velocity 150m per second from a pistol of mass 2 kg what is the recoil velocity of the pistol ?

\huge{\bold{\underline{\underline{....Answer....}}}}

....Answer....

\huge{\bold{\underline{Given:-}}}

Given:−

Mass of Bullet (m) = 20 grams.

Velocity of the Bullet after Firing (u) = 150 m/s.

Mass of The Pistol (M) = 2 kg.

Let the recoil velocity of the Pistol be "v".

\huge{\bold{\underline{Explanation:-}}}

Explanation:−

From Law of Conservation of momentum,

Initial Momentum = Final Momentum.

\large{\boxed{\tt P_i = P_f}}

P

i

=P

f

But Before Firing, the whole system (Gun & Bullet) is at Rest.

Therefore, Initial momentum will be zero.

Then the equation becomes,

\large{\leadsto {\underline{\underline{\tt P_f = 0}}}}⇝

P

f

=0

Now, Formulating the Values of Final Momentum.

\large{\tt \leadsto mu + Mv = 0}⇝mu+Mv=0

M = Mass of Pistol.

m = Mass of Bullet.

u = Velocity of bullet.

v = Velocity of bullet.

Substituting the values

\large{\tt \leadsto 20 \times 10^{-3} \times 150 + 2 \times v = 0}⇝20×10

−3

×150+2×v=0

[Here Mass of bullet (m) is taken as 20 grams = 20 × 10⁻³ Kg.]

\large{\tt \leadsto 20 \times 150 \times 10^{-3} + 2v = 0}⇝20×150×10

−3

+2v=0

\large{\tt \leadsto 3000 \times 10^{-3} + 2v = 0}⇝3000×10

−3

+2v=0

\large{\tt \leadsto 3\times 10^3 \times \dfrac{1}{10^3}+ 2v = 0}⇝3×10

3

×

10

3

1

+2v=0

\large{\tt \leadsto 3 \times \cancel{10^3} \times \dfrac{1}{\cancel{10^3}}+ 2v = 0}⇝3×

10

3

×

10

3

1

+2v=0

\large{\tt \leadsto 3 \times 1 + 2v = 0}⇝3×1+2v=0

\large{\tt \leadsto 3 + 2v = 0}⇝3+2v=0

\large{\tt \leadsto 2v = - 3}⇝2v=−3

\large{\tt \leadsto v = \dfrac{- 3}{2}}⇝v=

2

−3

\large{\tt \leadsto v = \cancel{\dfrac{- 3}{2}}}⇝v=

2

−3

\huge{\boxed{\boxed{\tt v = - 1.5 \: m/s}}}

v=−1.5m/s

So, The Recoil velocity of the Pistol is 1.5 m/s.

Note:-

Negative sign of Recoil velocity indicates that the recoil velocity of the Pistol is opposite to the Motion of the Bullet.

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