02: Physics And Mathematics : Cbse-xi-physics

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CBSE-XI-Physics

02: Physics and Mathematics

with Solutions - page 2

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#13-a
A→=B→,
Ans : `` \stackrel{\to }{A}=\stackrel{\to }{B},`` i.e., the two vectors are equal in magnitude and parallel to each other

#13-b
A→≠B→?
Ans : `` \stackrel{\to }{A}\ne \stackrel{\to }{B}``, i.e., the two vectors are unequal in magnitude and parallel or anti parallel to each other
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Qstn #14
Let
A→=5i→-4j→andB→=-7·5i→+6j→. Do we have
B→=kA→? Can we say
B→A→= k?
Ans : If `` \stackrel{\to }{A}=5\stackrel{\to }{i}-4\stackrel{\to }{j}\mathrm{and}\stackrel{\to }{B}=-7·5\stackrel{\to }{i}+6\stackrel{\to }{j}``, then we have `` \stackrel{\to }{B}=k\stackrel{\to }{A}`` by putting the value of scalar k as `` -1.5``.
However, we cannot say that `` \frac{\stackrel{\to }{B}}{\stackrel{\to }{A}}`` = k, because a vector cannot be divided by other vectors, as vector division is not possible.
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#
Section : B

Qstn #1
A vector is not changed if
-(a) it is rotated through an arbitrary angle
(b) it is multiplied by an arbitrary scalar
(c) it is cross multiplied by a unit vector
(d) it is slid parallel to itself.
digAnsr: d
Ans : (d) it is slid parallel to itself.
A vector is defined by its magnitude and direction. If we slide it to a parallel position to itself, then none of the given parameters, which define the vector, will change.
Let the magnitude of a displacement vector (`` \stackrel{\to }{A}``) directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.
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Qstn #2
Which of the sets given below may represent the magnitudes of three vectors adding to zero?
-(a) 2, 4, 8
(b) 4, 8, 16
(c) 1, 2, 1
(d) 0.5, 1, 2
digAnsr: c
Ans : (c) 1, 2, 1
1,2 and 1 may represent the magnitudes of three vectors adding to zero.For example one of the vector of length 1 should make an angle of `` {135}^{\circ }`` with x axis and the other vector of length 1 makes an angle of `` {225}^{\circ }`` with x axis. The third vector of length 2 should lie along x axis.
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Qstn #3
The resultant of
A→andB→makes an angle α with
A→and β with
B→,
-(a) α < β
(b) α < β if A < B
(c) α < β if A > B
(d) α < β if A = B
digAnsr: c
Ans : (c) α < β if A > B
The resultant of two vectors is closer to the vector with the greater magnitude.
Thus, α < β if A > B
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Qstn #4
The component of a vector is
-(a) always less than its magnitude
(b) always greater than its magnitude
(c) always equal to its magnitude
(d) None of these.
digAnsr: d
Ans : (d) None of these.
All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components.
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Qstn #5
A vector
A→points vertically upward and
B→points towards the north. The vector product
A→×B→is
-(a) along the west
(b) along the east
(c) zero
(d) vertically downward.
digAnsr: a
Ans : (a) along the west
The vector product `` \stackrel{\to }{A}\times \stackrel{\to }{B}`` will point towards the west. We can determine this direction using the right hand thumb rule.
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Qstn #6
The radius of a circle is stated as 2.12 cm. Its area should be written as
-(a) 14 cm²
(b) 14.1 cm²
(c) 14.11 cm²
(d) 14.1124 cm²
digAnsr: b
Ans : (b) 14.1 cm²
Area of a circle, A = `` \pi {r}^{2}``
On putting the values, we get:
`` A=\frac{22}{7}\times 2.12\times 2.12
\Rightarrow A=14.1{\mathrm{cm}}^{2}``
The rules to determine the number of significant digits says that in the multiplication of two or more numbers, the number of significant digits in the answer should be equal to that of the number with the minimum number of significant digits. Here, 2.12 cm has a minimum of three significant digits. So, the answer must be written in three significant digits.

#
Section : C

Qstn #1
A situation may be described by using different sets coordinate axes having different orientation. Which the following do not depended on the orientation of the axis?
.(a) the value of a scalar
(b) component of a vector
(c) a vector
(d) the magnitude of a vector.
digAnsr: a,c,d
Ans : (a) the value of a scalar
(c) a vector
(d) the magnitude of a vector
The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation. However, components of a vector depend on the orientation of the axes.
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Qstn #2
Let
C→=A→+B→(a)
C→is always greater than
A→(b) It is possible to have
C→<
A→and
C→<
B→(c) C is always equal to A + B
(d) C is never equal to A + B.
digAnsr: b
Ans : (b) It is possible to have `` \left|\stackrel{\to }{C}\right|`` < `` \left|\stackrel{\to }{A}\right|`` and `` \left|\stackrel{\to }{C}\right|`` < `` \left|\stackrel{\to }{B}\right|``
Statements (a), (c) and (d) are incorrect.
Given: `` \stackrel{\to }{C}=\stackrel{\to }{A}+\stackrel{\to }{B}``
Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of `` \stackrel{\to }{A}`` and `` \stackrel{\to }{B}`` or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.
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Qstn #3
Let the angle between two nonzero vectors
A→and B→be 120° and its resultant be
C→.
-(a) C must be equal to
A-B
(b) C must be less than
A-B
(c)
C must be greater than
A-B
(d) C may be equal to
A-B
digAnsr: c
Ans : (b) C must be less than `` \left|A-B\right|``
Here, we have three vector A, B and C.
`` {\left|\stackrel{\to }{A}+\stackrel{\to }{B}\right|}^{2}={\left|\stackrel{\to }{A}\right|}^{2}+{\left|\stackrel{\to }{B}\right|}^{2}+2\stackrel{\to }{A}.\stackrel{\to }{B}...\left(i\right)
{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}={\left|\stackrel{\to }{A}\right|}^{2}+{\left|\stackrel{\to }{B}\right|}^{2}-2\stackrel{\to }{A}.\stackrel{\to }{B}...\left(ii\right)``
Subtracting (i) from (ii), we get:
`` {\left|\stackrel{\to }{A}+\stackrel{\to }{B}\right|}^{2}-{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}=4\stackrel{\to }{A}.\stackrel{\to }{B}``
Using the resultant property `` \stackrel{\to }{C}=\stackrel{\to }{A}+\stackrel{\to }{B}``, we get:
`` {\left|\stackrel{\to }{C}\right|}^{2}-{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}=4\stackrel{\to }{A}.\stackrel{\to }{B}
\Rightarrow {\left|\stackrel{\to }{C}\right|}^{2}={\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}+4\stackrel{\to }{A}.\stackrel{\to }{B}
\Rightarrow {\left|\stackrel{\to }{C}\right|}^{2}={\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}+4\left|\stackrel{\to }{A}\right|\left|\stackrel{\to }{B}\right|\mathrm{cos}120°``
Since cosine is negative in the second quadrant, `` C``must be less than `` \left|A-B\right|``.

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Qstn #4
The x-component of the resultant of several vectors
-(a) is equal to the sum of the x-components of the vectors of the vectors
(b) may be smaller than the sum of the magnitudes of the vectors
(c) may be greater than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.
digAnsr: a,b,d
Ans : (a) is equal to the sum of the x-components of the vectors
(b) may be smaller than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.
The x-component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.
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