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A constant voltage is applied between the two ends of a uniform metallic wire, heat H is developed in it. If another wire of the same material, double the radius and twice the length as compared to original wire is used then the heat developed in it will be :
The heat developed in a wire is given by H = V²t / R. Resistance R is given by R = ρL/A.For the new wire: Length (L') = 2L, Radius (r') = 2r, so A' = 4A, R' = (ρ × 2L) / (4A) = R/2. Since H ∝ 1/R, the heat developed H' = 2H. Thus, the correct answer is (c) 2H. For more visit here: https://www.tiwariRead more
The heat developed in a wire is given by H = V²t / R. Resistance R is given by R = ρL/A.For the new wire:
Length (L’) = 2L,
Radius (r’) = 2r, so A’ = 4A,
R’ = (ρ × 2L) / (4A) = R/2.
Since H ∝ 1/R, the heat developed H’ = 2H.
Thus, the correct answer is (c) 2H.
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The current flowing through wire depends on time as, I = 3r + 2 t + 5. The charge flowing through the cross-section of wire in time t = 0 to t = 2 second is
To find the charge Q, integrate current I = 3t² + 2t + 5 over t = 0 to t =2: Q = ∫02(3t² +2t + 5)dt Solving, Q = t + t² + 5t 02 = (8+4+10) − (0) = 22C Answer: (a) 22 C. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
To find the charge
Q, integrate current I = 3t² + 2t + 5 over t = 0 to t =2:
Q = ∫02(3t² +2t + 5)dt
Solving,
Q = t + t² + 5t 02
= (8+4+10) − (0) = 22C
Answer: (a) 22 C.
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The mean free path of electrons in a metal is 4 X 10⁻⁸ m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in the units V/m,
The energy gained by an electron in an electric field E over a mean free path λ is given by: eEλ=2 eV Substituting e = 1.6 × 10⁻¹⁹ C and λ = 4 × 10 m: E = 2 1.6 × 10⁻¹⁹/4 × 10⁻⁸ E = 8 × 10⁷ V/m Thus, the correct answer is (d) 8 × 10⁷ V/m. For more visit here: https://www.tiwariacademy.com/ncert-soluRead more
The energy gained by an electron in an electric field E over a mean free path λ is given
by:
eEλ=2 eV
Substituting e = 1.6 × 10⁻¹⁹ C and
λ = 4 × 10 m:
E = 2 1.6 × 10⁻¹⁹/4 × 10⁻⁸
E = 8 × 10⁷ V/m
Thus, the correct answer is (d) 8 × 10⁷ V/m.
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A uniform copper wire of length 1 m and cross-sectional area 0-5 mm² carries a current of I A. Assuming that there are 8 x 10²² free electrons per cm³ in copper, how long will an electron take to drift from one end of the wire to the other end?
Drift velocity is vd = I/nAe , where n = 8 × 10²⁸ electrons/m³, A = 0.5 × 10⁻⁶ m², e = 1.6 × 10⁻¹⁹ C. Time t = L/vd gives 6.4 × 10³ s. Answer: (c) 6.4 × 10³ s. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
Drift velocity is
vd = I/nAe , where n = 8 × 10²⁸
electrons/m³, A = 0.5 × 10⁻⁶ m²,
e = 1.6 × 10⁻¹⁹ C. Time t =
L/vd gives 6.4 × 10³ s. Answer: (c) 6.4 × 10³ s.
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How does descriptive writing contribute to better lyric creation?
Descriptive writing plays a crucial role in lyric creation by adding vivid imagery, making songs more expressive and impactful. By using sensory details, songwriters can paint pictures with words, helping listeners visualize and feel the song’s emotions. Descriptive lyrics enhance storytelling, makiRead more
Descriptive writing plays a crucial role in lyric creation by adding vivid imagery, making songs more expressive and impactful. By using sensory details, songwriters can paint pictures with words, helping listeners visualize and feel the song’s emotions. Descriptive lyrics enhance storytelling, making the song more engaging and memorable. Whether describing a scene, an emotion or a concept, strong descriptive writing ensures that the song resonates deeply with the audience, increasing its emotional and artistic value.
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