In the figure shown, what is the current (in Ampere) drawn from the battery ?
You are given: R₁ = 15$$\Omega $$, R₂ = 10 $$\Omega $$, R₃ = 20 $$\Omega $$, R₄ = 5$$\Omega $$, R₅ = 25$$\Omega $$, R₆ = 30 $$\Omega $$, E = 15 V

A body of mass m₁ moving with an unknown velocity of $${v_1}\mathop i\limits^ \wedge $$, undergoes a collinear collision with a body of mass m₂ moving with a velocity $${v_2}\mathop i\limits^ \wedge $$ . After collision, m₁ and m₂ move with velocities of $${v_3}\mathop i\limits^ \wedge $$ and $${v_4}\mathop i\limits^ \wedge $$ , respectively. If m₂ = 0.5 m₁ and v₃ = 0.5 v₁, then v₁ is :-

A nucleus A, with a finite de-broglie wavelength $$\lambda $$_A, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths $$\lambda $$_B and $$\lambda $$_C of B and C are respectively :

Let $$\left| {\mathop {{A_1}}\limits^ \to } \right| = 3$$, $$\left| {\mathop {{A_2}}\limits^ \to } \right| = 5$$ and $$\left| {\mathop {{A_1}}\limits^ \to + \mathop {{A_2}}\limits^ \to } \right| = 5$$. The value of $$\left( {2\mathop {{A_1}}\limits^ \to + 3\mathop {{A_2}}\limits^ \to } \right)\left( {3\mathop {{A_1}}\limits^ \to - \mathop {2{A_2}}\limits^ \to } \right)$$ is :-