Select me MOST UPPropie upuUI TIUM USE YiVeil DCIU" 1. While performing experiment Ram observed that when UV-radiation were allowed to fall on the emitter plate of an evacuated glass tube enclosing two electrodes, current flow in the circuit due to (a)UV-light ionises the traces of gas left after evacuation (b) UV-light produces ionisation of cathode and anode (c) UV-light causes ionisation of cathode (d) UV-light causes ejection of electron from the emitter plate
Answers
Explanation:
Given :-
A bullet moving with a velocity of 50√2 m/s is fired into a target which it penetrates to the extent of d metres.
This bullet is fired into a target of d/2 metre thickness with the same velocity.
To Find :-
The bullet will come out of the target with a velocity.
Formula Used :-
\clubsuit♣ Third Equation Of Motion Formula :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}\\\end{gathered}
↦
v
2
=u
2
+2as
where,
v = Final Velocity
u = Initial Velocity
a = Acceleration
s = Distance Covered
Solution :-
\begin{gathered}{\small{\bold{\purple{\underline{\bigstar\: In\: first\: case\: :-}}}}}\\\end{gathered}
★Infirstcase:−
\leadsto⇝ A bullet moving with a velocity of 50√2 m/s is fired into a target which it penetrates to the extent of d metres.
Given :
Final Velocity = 0 m/s
Initial Velocity = 50√2 m/s
Distance Covered = d metres
According to the question by using the formula we get,
\implies \sf (0)^2 =\: (50\sqrt{2})^2 + 2ad⟹(0)
2
=(50
2
)
2
+2ad
\implies \sf 0 \times 0 =\: 50\sqrt{2} \times 50\sqrt{2} + 2ad⟹0×0=50
2
×50
2
+2ad
\implies \sf 0 =\: 2500 \times 2 + 2ad⟹0=2500×2+2ad
\implies \sf 0 =\: 5000 + 2ad⟹0=5000+2ad
\implies \sf 0 - 5000 =\: 2ad⟹0−5000=2ad
\implies \sf - 5000 =\: 2ad⟹−5000=2ad
\implies \sf \dfrac{- \cancel{5000}}{\cancel{2}d} =\: a⟹
2
d
−
5000
=a
\implies \sf \dfrac{- 2500}{d} =\: a⟹
d
−2500
=a
\begin{gathered}\implies \sf\bold{\green{a =\: \dfrac{- 2500}{d}}}\\\end{gathered}
⟹a=
d
−2500
Hence, the acceleration is - 2500/d .
\begin{gathered}{\small{\bold{\purple{\underline{\bigstar\: In\: second\: case\: :-}}}}}\\\end{gathered}
★Insecondcase:−
\leadsto⇝ This bullet is fired into a target of d/2 metre thickness with the same velocity.
Given :
Acceleration = - 2500/d
Initial Velocity = 50√2 m/s
Distance Covered = d/2 metre
According to the question by using the formula we get,
\longrightarrow \sf v^2 =\: (50\sqrt{2})^2 + 2 \times \dfrac{- 2500}{d} \times \dfrac{d}{2}⟶v
2
=(50
2
)
2
+2×
d
−2500
×
2
d
\longrightarrow \sf v^2 =\: 50\sqrt{2} \times 50\sqrt{2} + \dfrac{- 5000}{d} \times \dfrac{d}{2}⟶v
2
=50
2
×50
2
+
d
−5000
×
2
d
\longrightarrow \sf v^2 =\: 2500 \times 2 + \dfrac{- 5000}{\cancel{d}} \times \dfrac{\cancel{d}}{2}⟶v
2
=2500×2+
d
−5000
×
2
d
\longrightarrow \sf v^2 =\: 5000 + \dfrac{- 5000}{2}⟶v
2
=5000+
2
−5000
\longrightarrow \sf v^2 =\: 5000 + (- 2500)⟶v
2
=5000+(−2500)
\longrightarrow \sf v^2 =\: 5000 - 2500⟶v
2
=5000−2500
\longrightarrow \sf v^2 =\: 2500⟶v
2
=2500
\longrightarrow \sf v =\: \sqrt{2500}⟶v=
2500
\longrightarrow \sf\bold{\red{v =\: 50\: m/s}}⟶v=50m/s
\therefore∴ The bullet will come out of the target with a velocity is 50 m/s .
Hence, the correct options is option no (c) 50 m/s.