UGEE SUPR Mock Test- 6 - JEE MCQ

g = 10 m/s²

By conservation of energy, Potential energy lost by the solid sphere in rolling down the inclined plane = Kinetic energy gained by the sphere.

P = ?_o?E

where, E is the electric field, and x is electric susceptibility, and it is a dimensionless quantity, Dimensional formula of P = Dimensional formula of (ε₀) x Dimensional formula of E

= [M^-1 L^-3 T⁴ I²] [ML T^-3I^-1] = [M⁰ L^-2 T^-1 I¹]

and

i.e., r₂ = 2r₁

Since, volume V ∝ r³

∴ V₂/V₁ = (r₂/r₁)³ = (2r₁/r₁)³ = 8

⇒ V₂ = 8V₁ ….(i)

From Boyle's law,

p₁V₁ = p₂V₂

Where, p₂ = atmospheric pressure.

p₁V₁ = p₂ 8V₁

(from Eq. (i)J

p₂ = p₁/8

Where. p₂ = atmospheric pressure and p₁ = pressure at depth d.

∴ p₁ = p_atmospheric + ρ gd

P₁ = p₂ + ρ gd

From Eqs. (ii) and (iii), we have

p₂ = p₂ + ?gd

7p₂ = ρ gd

7.ρgH = ρ gd

7H = d ⇒ d = 7 H

Hence, V_A = V_B

On replacing a new cell of emf 2E and internal resistance 3r, no effect occurs on balanced condition of bridge, i.e., V_A = V_B.

Hence, the galvanometer wil shows zero deflection.

Torque = Force x Perpendicular distance

i.e., τ = F x r

∴ Dimensional formula of τ = [M L T^-2] [L] = [ML² T^-2]

E_i = 2π² mr²₁f²_{1 ………..}(i)

Where, f₁ = frequency of revolution of the body.

When, r₂ = 2r₁ and f₂ = and 2f₁ , then

Change in kinetic energy,

ΔE = E_r -E_i

= 16E_i - E_i = 15 E_i

i.e …(i)

where, T = temperature and M = molecular mass. When the temperature is doubled, then O₂ dissociates into two atoms.

Hence, the atomic mass of the oxygen atom.

(M') = M/2

∴ rms velocity of the oxygen atom =

When 8 small drops are combined to form a bigger drop, then volume will be same.

If r₁ be radius of small drop and r₂ be the radius of biggordrop, then v₂ = v₁

∴ Terminal velocity of the drop

i.e., v ∝ r²

⇒ v₁/v₂ = 1/4

v₂ = 4v₁ x 10 = 40 cm/s

Hence, work done by the centripetal force is zero, when it through (2/3)rd of the circular path.

Magnetic flux, ϕ = 4 x 10^-5 Wb

Cross-sectional area, A = 0.4 cm² = 4 x 10^-5 m²

∴ Permeability of rod.

∴ F = (M + m)a

⇒ a = F/M+m

E = (1/2) mω²A²

Where, m = mass of body performing SHM.

ω = angular velocity

and A = amplitude.

E ∝ A²

(a)

(b)

Other isomers are not primary amines.

In the given complex ion [Fe(CN)₆]^3-, the Fe is in +3 oxidation state. As we know that atomic number of iron is 26. The number of electron in Fe³⁺ is 23. Each of the six cyanide molecules donates a pair of elections so that EAN becomes 23 + 2 x 6 ⇒ 35.

At constant pressure, q_p = ΔU -W.

q_p = ΔU - pΔV

q_p = U₂ - U₁ - (-p(V₂ - V₁))

q_p = (U₂ + pV₂) - (U₁ + pV₁)

q_p = H₂ - H₁

q_p = ΔH

As we know, ΔH = ΔU + pΔV

ΔH = ΔU + W

Putting the value of ΔH in (1)

q_p = ΔU + W

Other corrected options are as follows:

• For adiabatic process ΔU = W

• For an isochoric process ΔU = q_v

• For an isothermal process q = -W

Electrochemical equivalent = Equivalent weight/96500

gm per coulomb. Its unit is kgC^-1.