PHYSICS MOCK JEE ADVANCED ( answers), Exams of Physics

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PHYSICS MOCK

JEE ADVANCED

• 𝑇ℎ𝑖𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑭𝑶𝑼𝑹 (𝟒) 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠.
• 𝐸𝑎𝑐ℎ 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛 ℎ𝑎𝑠 𝑓𝑜𝑢𝑟 𝑜𝑝𝑡𝑖𝑜𝑛𝑠, (𝐴), (𝐵), (𝐶) 𝑎𝑛𝑑 (𝐷). 𝑶𝑵𝑳𝒀 𝑶𝑵𝑬 𝑜𝑓 𝑡ℎ𝑒𝑠𝑒 4

• 𝑀𝑎𝑟𝑘𝑖𝑛𝑔 𝑠𝑐ℎ𝑒𝑚𝑒: + 3 𝑓𝑜𝑟 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑐ℎ𝑜𝑖𝑐𝑒

1

2

1

2

. [𝑇𝑎𝑘𝑒 𝐹 =

]

(A)

12 𝑀𝑔

3

(B)

13 𝑀𝑔

6

(C)

17 𝑀𝑔

6

(D)

19 𝑀𝑔

3

(A)

5 𝜆

2

2 𝜖

𝑜

(B)

3 𝜆

2

2 𝜖

𝑜

(C)

𝜆

2

2 𝜖

𝑜

(D)

𝜆

2

𝜖

𝑜

3

2

(A) 𝐼𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑖𝑛𝑎𝑙 𝑖𝑚𝑎𝑔𝑒 𝑓𝑜𝑟 𝑎𝑛 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟, 𝑙𝑒𝑓𝑡 𝑜𝑓 ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑒

(B) 𝐼𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑖𝑛𝑎𝑙 𝑖𝑚𝑎𝑔𝑒 𝑓𝑜𝑟 𝑎𝑛 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟, 𝑙𝑒𝑓𝑡 𝑜𝑓 ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑒

(C) 𝐴𝑛 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟 𝑖𝑠 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑠𝑝𝑒𝑐𝑖𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑔𝑙𝑎𝑠𝑠 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

(D) 𝐴𝑛 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟 𝑖𝑠 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑠𝑝𝑒𝑐𝑖𝑎𝑙 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑔𝑙𝑎𝑠𝑠 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

− 1

− 1

− 1

− 1

𝐴

𝐵

(A) 16 (B) 4 (C)

1

4

(D)

1

16

1

2

(A) 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝐿

1

1

(B) 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝐿

2

2

(C) 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝐿

2

(D) 𝑇𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑜𝑠𝑐𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐ℎ𝑎𝑟𝑔𝑒 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑤ℎ𝑒𝑛 𝑏𝑜𝑡ℎ 𝑘𝑒𝑦𝑠 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑.

48 𝑉. 𝑇ℎ𝑒 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎 𝑎𝑛𝑑 𝑏 𝑖𝑠 ___(in V)

𝑜

sin

2

𝑜

− 2

𝑜

− 2

√ 3 × 10

− 2

%. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑛 𝑖𝑠 __

𝑜

𝑜

𝒐

𝒐

𝑅

2

𝟒

𝐹𝑖𝑙𝑙 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑁: ___

1

2

𝑎 × 𝑑

𝑏 × 𝛽(

1

2

, 𝑡ℎ𝑒𝑛 𝑎 + 𝑏 = ___

Paragraph for Question 16 and 17

1

16. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑠 ____

17. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 |𝑦| 𝑖𝑠 ____

Q1: B Q2: C Q3: C Q4: C Q5: AD Q6: A Q7: BCD

Q8: 91 Q9: 204 Q10: 4 Q11: 3 Q12: 21 Q13: 3 Q14: 4

Q15: 1 Q16: 140 Q17: 0.

Q1:

Block A (x-axis): 2 𝐹 − 𝑁

𝐴𝐶

sin

∘

𝐴𝐶

𝐴𝐶

Block A (y-axis): 𝑁

1

𝐴𝐶

cos( 30

∘

Block B (x-axis): 𝑁

𝐶𝐵

sin( 30

∘

𝐶𝐵

𝐶𝐵

Block B (y-axis): 𝑁

2

𝐶𝐵

cos

∘

2

1

2

Substitute 𝐹 =

1

2

Q2:

Field from Wire 1 at distance 𝑟 =

2

2

is 𝐸 =

0

By symmetry, only the force component along the shortest distance (y-axis) survives:

𝑦

𝐸 cos 𝜃 =

0

2

2

2

2

𝑦

2

0

2

2

𝑦

∞

−∞

2

0

[

tan

− 1

)]

−∞

∞

2

0

2

0

1

cm

2

(− 2 + 2 ) = 0 cm ⇒ Δ𝑥 =

cm (Option D)

Q6:

Rate of heat flow 𝑄 =

𝑡ℎ

where thermal resistance 𝑅

𝑡ℎ

For Window A (4mm glass): 𝑅

𝐴

4 × 10

− 3

5 × 10

− 3

For Window B (2mm air gap, thin glass ignored): 𝑅 𝐵

2 × 10

− 3

80 × 10

− 3

𝐴

𝐵

𝐵

𝐴

80 × 10

− 3

5 × 10

− 3

Q7:

𝑚𝑎𝑥

0

2

( 10 mC)

2

40 mF

= 1. 25 mJ

(A) 𝑘

1

only: 𝐼

1

𝑚𝑎𝑥

1

= 10 mA (False)

(B) 𝑘

2

only: 𝐼

2

𝑚𝑎𝑥

2

= 5 mA (True)

(C) Both closed (parallel): 𝐿

1

1

2

2

1

2

1

2

𝑚𝑎𝑥

1

1

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

= √ 0. 005 A = √ 5 mA (True)

Q 8 :

Continuous path implies alternating directions: odd semicircles expand

(+𝑊), even compress (−𝑊).

𝐵

𝐵

𝐴

𝐴

0

0

0

0

0

0

Net linear work (rectangular bases + isobaric connectors) 𝑊

𝑙𝑖𝑛

0

𝐵

𝐴

0

0

Semicircular work 𝑊

𝑠𝑒𝑚𝑖

1

2

3

0

0

𝑛

∞

𝑛= 0

𝑠𝑒𝑚𝑖

0

0

0

0

0

0

𝑙𝑖𝑛

𝑠𝑒𝑚𝑖

0

0

0

0

0

0

0

0

0

0

0

0

1

2

1

2

Q 9 :

Upper block

1

12 × 6

Total current 𝐼 =

1

1

= 12 A

Middle block ( 30 Ω ∥ 15 Ω ∥ 10 Ω):

2

2

Total equivalent resistance: 𝑅

𝑒𝑞

1

2

Total potential difference 𝑉

𝑎𝑏

= 𝐼 × 𝑅

𝑒𝑞

= 12 × 17 = 204 V

Q10:

0

sin

2

0

sin 𝜃 cos 𝜃 ⋅ Δ𝜃 = 𝐼

0

sin( 2 𝜃) Δ

1. 0020 = 20 sin( 60

∘

) Δ𝜃 ⇒ 20 × 10

− 4

10 √ 3 Δ𝜃 = 20 × 10

− 4

× 10

− 4

rad

Q13:

Temperature gradient

2

1

2

1

Substitute 𝑑𝑥: 𝑑𝑡 =

2

1

2

1

𝑇

2

𝑇 1

− 1 / 2

2

1

[ 2 √𝑇]

𝑇

1

𝑇

2

2

1

2

1

Factor the denominator: 𝑇

2

1

2

1

2

1

2

1

2

1

2

1

1

2

a + b = 3

Q14:

Since ℎ ≪ 𝑅, the gravitational acceleration is nearly constant outside the shell:

′

𝑠𝑝ℎ𝑒𝑟𝑒

𝑠ℎ𝑒𝑙𝑙

2

2

2

Using 1D kinematics: ℎ =

′

2

′

2

2

Q15:

Apply conservation of energy. Since ℎ ≪ 𝑅, approximate initial distance as 2 𝑅.

Initial Potential Energy: 𝑈

𝑖

Final Potential Energy at B (inside shell, on sphere): 𝑈

𝑓

𝑠ℎ𝑒𝑙𝑙

𝑠𝑝ℎ𝑒𝑟𝑒

𝑓

𝑖

𝑖

𝑓

𝑓

2

2

2

Q16: &

For Lens 1: 𝑢

1

= − 20 cm, 𝑓

1

= + 15 cm

1

1

1

= 60 cm

Magnification 𝑚

1

1

1

y-coordinate of tip after 𝐿

1

1

1

× ℎ

𝑜

= − 3 × 0. 2 cm = − 0. 6 cm

Mirror is at 𝑦 = − 0. 4 cm. The tip is formed below it, so it reflects:

Reflected tip 𝑦

1

′

𝑚𝑖𝑟𝑟𝑜𝑟

𝑚𝑖𝑟𝑟𝑜𝑟

1

) = − 0. 4 + (− 0. 4 − (− 0. 6 )) = − 0. 2 cm

For Lens 2: The object is the reflected image at 𝑥 = 60 cm

2

= −( 120 − 60 ) = − 60 cm, 𝑓

2

= + 15 cm

2

2

2

= 20 cm

Final x-coordinate 𝑥 = 120 + 20 = 140 cm (Answer 16)

Magnification 𝑚

2

2

2

Final y-coordinate 𝑦 = 𝑚

2

× 𝑦

1

′

×

− 0. 2 cm

cm =

cm