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Work done in carrying an electron from AxoB lying on an equipotential surface of one volt potential is
The correct answer is (d) Zero. On an equipotential surface, all points have the same electric potential. Since work done (W = qΔV) depends on the potential difference (ΔV) and ΔV = 0 on an equipotential surface, no work is required to move an electron between points. For more visit here: https://wwRead more
The correct answer is (d) Zero.
On an equipotential surface, all points have the same electric potential. Since work done (W = qΔV) depends on the potential difference (ΔV) and ΔV = 0 on an equipotential surface, no work is required to move an electron between points.
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A charge of 10 p. C lies at the centre of a square. Work done in carrying a charge of 2 p C from one comer of square to the diagonally opposite comer is
The correct answer is (c) Zero. The work done in moving a charge in an electric field is given by W = qΔV. Since the charge is at the center of the square, all corners are at the same potential. The potential difference (ΔV) between diagonally opposite corners is zero, making the work done also zeroRead more
The correct answer is (c) Zero.
The work done in moving a charge in an electric field is given by W = qΔV. Since the charge is at the center of the square, all corners are at the same potential. The potential difference (ΔV) between diagonally opposite corners is zero, making the work done also zero, regardless of the path taken.
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N/C is the SI unit of
(c) electric intensity The SI unit of electric field intensity (also called electric field strength) is newtons per coulomb (N/C). This represents the force per unit charge exerted on a small positive test charge placed in the field. For more visit here: https://www.tiwariacademy.com/ncert-solutionsRead more
(c) electric intensity
The SI unit of electric field intensity (also called electric field strength) is newtons per coulomb (N/C). This represents the force per unit charge exerted on a small positive test charge placed in the field.
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If 8 tan x = 15, then sin x – cos x is equal to
We are given: 8 tan x = 15. Step 1: Solve for tan x Rearrange the equation to solve for tan x: tan x = 15/8. Step 2: Express sin x and cos x in terms of tan x Using the identity tan x = sin x / cos x, we can write: sin x = 15k and cos x = 8k, where k is a positive constant such that sin²x + cos²x =Read more
We are given:
8 tan x = 15.
Step 1: Solve for tan x
Rearrange the equation to solve for tan x:
tan x = 15/8.
Step 2: Express sin x and cos x in terms of tan x
Using the identity tan x = sin x / cos x, we can write:
sin x = 15k and cos x = 8k,
where k is a positive constant such that sin²x + cos²x = 1 (Pythagorean identity).
Substitute sin x = 15k and cos x = 8k into the identity:
(15k)² + (8k)² = 1
225k² + 64k² = 1
289k² = 1
k² = 1/289
k = √(1/289)
k = 1/17.
Thus:
sin x = 15k = 15/17,
cos x = 8k = 8/17.
Step 3: Find sin x – cos x
Now, calculate sin x – cos x:
sin x – cos x = (15/17) – (8/17)
= (15 – 8)/17
= 7/17.
Step 4: Final Answer
The value of sin x – cos x is 7/17.
The correct answer is:
d) 7/17
This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.
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The distance of the point P(2, 3) from the x- axis is
Finding the Distance of a Point from the x-Axis We are given a point P(2,3) and need to find its distance from the x-axis. Step 1: Understanding Distance from the x-Axis The x-axis is the horizontal axis in the Cartesian plane. The distance of any point (x, y) from the x-axis is simply the absoluteRead more
Finding the Distance of a Point from the x-Axis
We are given a point P(2,3) and need to find its distance from the x-axis.
Step 1: Understanding Distance from the x-Axis
The x-axis is the horizontal axis in the Cartesian plane.
The distance of any point (x, y) from the x-axis is simply the absolute value of its y-coordinate.
This is because the x-axis is located at y=0, and the vertical distance from any point to this axis is determined by how far its y-value is from zero.
Step 2: Applying the Formula
For a point (x,y), the distance from the x-axis is given by:
Distance = |y|
For point P(2, 3), we substitute y = 3:
Distance = |3| = 3
Step 3: Final Answer
Thus, the distance of point P(2,3) from the x-axis is 3.
Correct Option: (b) 3
This question related to Chapter 7 Mathematics Class 10th NCERT. From the Chapter 7 Coordinate Geometry. Give answer according to your understanding.
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