(a) Increase in temperature of water. Evaporation increases with temperature because higher thermal energy allows more water molecules to escape into the air. Increased kinetic energy breaks intermolecular forces, leading to faster evaporation. Other factors, like reduced surface area or adding saltRead more
(a) Increase in temperature of water. Evaporation increases with temperature because higher thermal energy allows more water molecules to escape into the air. Increased kinetic energy breaks intermolecular forces, leading to faster evaporation. Other factors, like reduced surface area or adding salt, decrease evaporation.
The correct answer is (b) 2.5 × 10¹⁸. Using Ohm’s law, I = V/R =200V/100Ω =2A. The charge flow per second is Q = It = 2C. The number of electrons is Q/e = 2/(1.6 × 10⁻¹⁹) = 2.5 ×10¹⁸. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-1/
The correct answer is (b) 2.5 × 10¹⁸. Using Ohm’s law,
I = V/R =200V/100Ω =2A. The charge flow per second is
Q = It = 2C. The number of electrons is
Q/e = 2/(1.6 × 10⁻¹⁹) = 2.5 ×10¹⁸.
(d) Zero. The electric potential due to a dipole at any point on its equatorial line is always zero. This is because the contributions from the positive and negative charges cancel each other out, as they are equidistant from the point and their potentials have equal magnitudes but opposite signs, rRead more
(d) Zero. The electric potential due to a dipole at any point on its equatorial line is always zero. This is because the contributions from the positive and negative charges cancel each other out, as they are equidistant from the point and their potentials have equal magnitudes but opposite signs, resulting in a net potential of zero.
The correct answer is (d) +5e, -8e, +7e. When conductors come in contact, charge redistributes while conserving total charge. The initial total charge is (+3e) + (+5e) + (-3e) = +5e. Among the given options, only (d) +5e, -8e, +7e maintains this total charge, satisfying charge conservation and possiRead more
The correct answer is (d) +5e, -8e, +7e. When conductors come in contact, charge redistributes while conserving total charge. The initial total charge is (+3e) + (+5e) + (-3e) = +5e. Among the given options, only (d) +5e, -8e, +7e maintains this total charge, satisfying charge conservation and possible redistribution after separation.
(b) 35.5 u. Chlorine has two isotopes, ³⁵Cl (75%) and ³⁷Cl (25%), and its atomic mass is the weighted average of these isotopes. The calculation gives (35 × 0.75) + (37 × 0.25) = 35.5 u, making 35.5 u the correct atomic mass. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
(b) 35.5 u. Chlorine has two isotopes, ³⁵Cl (75%) and ³⁷Cl (25%), and its atomic mass is the weighted average of these isotopes. The calculation gives (35 × 0.75) + (37 × 0.25) = 35.5 u, making 35.5 u the correct atomic mass.
Which condition out of the following will increase the evaporation of water? (a) Increase in temperature of water (b) Decrease in temperature of water (c) Less exposed surface area of water (d) Adding common salt to water
(a) Increase in temperature of water. Evaporation increases with temperature because higher thermal energy allows more water molecules to escape into the air. Increased kinetic energy breaks intermolecular forces, leading to faster evaporation. Other factors, like reduced surface area or adding saltRead more
(a) Increase in temperature of water. Evaporation increases with temperature because higher thermal energy allows more water molecules to escape into the air. Increased kinetic energy breaks intermolecular forces, leading to faster evaporation. Other factors, like reduced surface area or adding salt, decrease evaporation.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-1/
See lessA potential difference of 200 V is maintained across a conductor of resistance 100 Ω The number of electrons passing through it in 1 second is
The correct answer is (b) 2.5 × 10¹⁸. Using Ohm’s law, I = V/R =200V/100Ω =2A. The charge flow per second is Q = It = 2C. The number of electrons is Q/e = 2/(1.6 × 10⁻¹⁹) = 2.5 ×10¹⁸. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-1/
The correct answer is (b) 2.5 × 10¹⁸. Using Ohm’s law,
I = V/R =200V/100Ω =2A. The charge flow per second is
Q = It = 2C. The number of electrons is
Q/e = 2/(1.6 × 10⁻¹⁹) = 2.5 ×10¹⁸.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-1/
The electric potential on the axis of an electric dipole at a distance r from its centre is V. Then the potential at a point at the same distance on itsequatorial line will be
(d) Zero. The electric potential due to a dipole at any point on its equatorial line is always zero. This is because the contributions from the positive and negative charges cancel each other out, as they are equidistant from the point and their potentials have equal magnitudes but opposite signs, rRead more
(d) Zero. The electric potential due to a dipole at any point on its equatorial line is always zero. This is because the contributions from the positive and negative charges cancel each other out, as they are equidistant from the point and their potentials have equal magnitudes but opposite signs, resulting in a net potential of zero.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-1/
In a chamber, three spheres carry charges + 3 e, + 5 e and – 3 e. They came in contact for a moment and got separated. Which one of the following are possible values for the final charges on the spheres?
The correct answer is (d) +5e, -8e, +7e. When conductors come in contact, charge redistributes while conserving total charge. The initial total charge is (+3e) + (+5e) + (-3e) = +5e. Among the given options, only (d) +5e, -8e, +7e maintains this total charge, satisfying charge conservation and possiRead more
The correct answer is (d) +5e, -8e, +7e. When conductors come in contact, charge redistributes while conserving total charge. The initial total charge is (+3e) + (+5e) + (-3e) = +5e. Among the given options, only (d) +5e, -8e, +7e maintains this total charge, satisfying charge conservation and possible redistribution after separation.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-1/
Atomic mass of chlorine is
(b) 35.5 u. Chlorine has two isotopes, ³⁵Cl (75%) and ³⁷Cl (25%), and its atomic mass is the weighted average of these isotopes. The calculation gives (35 × 0.75) + (37 × 0.25) = 35.5 u, making 35.5 u the correct atomic mass. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
(b) 35.5 u. Chlorine has two isotopes, ³⁵Cl (75%) and ³⁷Cl (25%), and its atomic mass is the weighted average of these isotopes. The calculation gives (35 × 0.75) + (37 × 0.25) = 35.5 u, making 35.5 u the correct atomic mass.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/
See less