
Chapter 2 – Physics and Mathematics
Page No 27:
Question 1:
Is a vector necessarily changed if it is rotated through an angle?
Answer:
Yes. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed.
Question 2:
Is it possible to add two vectors of unequal magnitudes and get zero? Is it possible to add three vectors of equal magnitudes and get zero?
Answer:
No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors
A→and
B→of unequal magnitudes acting in opposite directions. The resultant vector is given by
R=A2+B2+2ABcosθIf two vectors are exactly opposite to each other, then
θ=180°, cos180°=1R=A2+B22AB⇒R=AB2⇒R=AB or BAFrom the above equation, we can say that the resultant vector is zero (R = 0) when the magnitudes of the vectors
A→and
B→are equal (A = B) and both are acting in the opposite directions.
Yes, it is possible to add three vectors of equal magnitudes and get zero.
Lets take three vectors of equal magnitudes
A, →B→ and C→, given these three vectors make an angle of
120°with each other. Consider the figure below:
Lets examine the components of the three vectors.
Ax=AAy=0Bx=B cos 60°By=B sin 60°Cx=C cos 60°Cy=C sin 60°Here, A=B=CSo, along the xaxis , we have:A(2A cos 60°)=0, as cos 60°=12 ⇒B sin 60°C sin 60°=0Hence, proved.
Question 3:
Does the phrase “direction of zero vector” have physical significance? Discuss it terms of velocity, force etc.
Answer:
A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
For any vector
A→, assume that
A→+0→=A→A→0→=A→A→×0→=0→Again, for any real number
λ, we have:
λ0→=0→The significance of a zero vector can be better understood through the following examples:
The displacement vector of a stationary body for a time interval is a zero vector.
Similarly, the velocity vector of the stationary body is a zero vector.
When a ball, thrown upward from the ground, falls to the ground, the displacement vector is a zero vector, which defines the displacement of the ball.
Question 4:
Can you add three unit vectors to get a unit vector? Does your answer change if two unit vectors are along the coordinate axes?
Answer:
Yes we can add three unit vectors to get a unit vector.
No, the answer does not change if two unit vectors are along the coordinate axes. Assume three unit vectors
i,^i^ and j^along the positive xaxis, negative xaxis and positive yaxis, respectively. Consider the figure given below:
The magnitudes of the three unit vectors (
i,^i^ and j^) are the same, but their directions are different.
So, the resultant of
i^ and i^is a zero vector.
Now,
j^+0→=j^ (Using the property of zero vector)
∴ The resultant of three unit vectors (
i,^i^ and j^) is a unit vector (
j^).
Page No 28:
Question 5:
Can we have physical quantities having magnitude and direction which are not vectors?
Answer:
Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.
Question 6:
Which of the following two statements is more appropriate?
(a) Two forces are added using triangle rule because force is a vector quantity.
(b) Force is a vector quantity because two forces are added using triangle rule.
Answer:
Two forces are added using triangle rule, because force is a vector quantity. This statement is more appropriate, because we know that force is a vector quantity and only vectors are added using triangle rule.
Question 7:
Can you add two vectors representing physical quantities having different dimensions? Can you multiply two vectors representing physical quantities having different dimensions?
Answer:
No, we cannot add two vectors representing physical quantities of different dimensions. However, we can multiply two vectors representing physical quantities with different dimensions.
Example: Torque,
τ→=r→×F→
Question 8:
Can a vector have zero component along a line and still have nonzero magnitude?
Answer:
Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Example: Consider a two dimensional vector
2i^+0j^. This vector has zero components along a line lying along the Yaxis and a nonzero component along the Xaxis. The magnitude of the vector is also nonzero.
Now, magnitude of
2i^+0j^=
22+02=2
Question 9:
Let ε_{1} and ε_{2} be the angles made by
A→and
A→with the positive Xaxis. Show that tan ε_{1} = tan ε_{2}_{.} Thus, giving tan ε does not uniquely determine the direction of
A→.
Answer:
The direction of
A→is opposite to
A→. So, if vector
A→and
A→make the angles ε_{1} and ε_{2} with the Xaxis, respectively, then ε_{1} is equal to ε_{2} as shown in the figure:
Here, tan ε_{1} = tan ε_{2}
Because these are alternate angles.
Thus, giving tan ε does not uniquely determine the direction of
A→.
Question 10:
Is the vector sum of the unit vectors
i→and
j→a unit vector? If no, can you multiply this sum by a scalar number to get a unit vector?
Answer:
No, the vector sum of the unit vectors
i→and
j→is not a unit vector, because the magnitude of the resultant of
i→and
j→is not one.
Magnitude of the resultant vector is given by
R =
12+12+cos90°=2Yes, we can multiply this resultant vector by a scalar number
12to get a unit vector.
Question 11:
Let
A→=3i→+4j→. Write a vector
B→such that
A→≠B→, but A = B.
Answer:
A vector
B→such that
A→≠B→, but A = B are as follows:
(i) B→=3i→4j→(ii) B→=3j→+4k→(iii) B→=3k→+4i→(iv) B→=3j→4k→
Question 12:
Can you have
A→×B→=A→·B→with A ≠ 0 and B ≠ 0? What if one of the two vectors is zero?
Answer:
No, we cannot have
A→×B→=A→·B→with A ≠ 0 and B ≠ 0. This is because the left hand side of the given equation gives a vector quantity, while the right hand side gives a scalar quantity. However, if one of the two vectors is zero, then both the sides will be equal to zero and the relation will be valid.
Question 13:
If
A→×B→=0, can you say that (a)
A→=B→,(b)
A→≠B→?
Answer:
If
A→×B→=0, then both the vectors are either parallel or antiparallel, i.e., the angle between the vectors is either
0° or 180°.
(
A→B→sinθn^=0
∵ sin0°=sin180°=0)
Both the conditions can be satisfied:
(a)
A→=B→,i.e., the two vectors are equal in magnitude and parallel to each other
(b)
A→≠B→, i.e., the two vectors are unequal in magnitude and parallel or anti parallel to each other
Question 14:
Let
A→=5i→4j→ and B→=7·5i→+6j→. Do we have
B→=kA→? Can we say
B→A→= k?
Answer:
If
A→=5i→4j→ and B→=7·5i→+6j→, then we have
B→=kA→by putting the value of scalar k as
1.5.
However, we cannot say that
B→A→= k, because a vector cannot be divided by other vectors, as vector division is not possible.
Question 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.
Answer:
(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 (
A→) directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.
Question 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
Answer:
(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∘with x axis and the other vector of length 1 makes an angle of
225∘with x axis. The third vector of length 2 should lie along x axis.
Question 3:
The resultant of
A→ and B→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
Answer:
(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
Question 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.
Answer:
(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.
Question 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.
Answer:
(a) along the west
The vector product
A→×B→will point towards the west. We can determine this direction using the right hand thumb rule.
Question 6:
The radius of a circle is stated as 2.12 cm. Its area should be written as
(a) 14 cm^{2}
(b) 14.1 cm^{2}
(c) 14.11 cm^{2}
(d) 14.1124 cm^{2}
Answer:
(b) 14.1 cm^{2}
Area of a circle, A =
πr2On putting the values, we get:
A=227×2.12×2.12⇒A=14.1 cm2
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.
Question 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.
Answer:
(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.
Question 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.
Answer:
(b) It is possible to have
C→<
A→and
C→<
B→Statements (a), (c) and (d) are incorrect.
Given:
C→=A→+B→Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of
A→and
B→or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.
Question 3:
Let the angle between two nonzero vectors
A→and
B→be 120° and its resultant be
C→.
(a) C must be equal to
AB(b) C must be less than
AB(c) C must be greater than
AB(d) C may be equal to
AB
Answer:
(b) C must be less than
ABHere, we have three vector A, B and C.
A→+B→2=A→2+B→2+2A→.B→ …(i)A→B→2=A→2+B→22A→.B→ …(ii)Subtracting (i) from (ii), we get:
A→+B→2A→B→2=4A→.B→Using the resultant property
C→=A→+B→, we get:
C→2A→B→2=4A→.B→⇒C→2=A→B→2+4A→.B→⇒C→2=A→B→2+4A→B→cos120°Since cosine is negative in the second quadrant,
Cmust be less than
AB.
Question 4:
The xcomponent of the resultant of several vectors
(a) is equal to the sum of the xcomponents 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.
Answer:
(a) is equal to the sum of the xcomponents 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 xcomponent of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.
Question 5:
The magnitude of the vector product of two vectors
A→and
B→may be
(a) greater than AB
(b) equal to AB
(c) less than AB
(d) equal to zero.
Answer:
(b) equal to AB
(c) less than AB
(d) equal to zero.
The magnitude of the vector product of two vectors
A→and
B→may be less than or equal to AB, or equal to zero, but cannot be greater than AB.
Page No 29:
Question 1:
A vector
A→makes an angle of 20° and
B→makes an angle of 110° with the Xaxis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.
Answer:
From the above figure, we have:
Angle between
A→and
B→= 110° − 20° = 90°
A→=3 m and B→= 4 mMagnitude of the resultant vector is given by
R=A2+B2+2AB cos θ =32+42+2×3×4×cos 90° = 5 mLet β be the angle between
R→ and A→.
β=tan1 A sin 90°A+B cos 90° =tan1 4 sin 90°3+4 cos 90° =tan1 43 =tan1 1.333 =53°Now, angle made by the resultant vector with the Xaxis = 53° + 20° = 73°
∴ The resultant
R→is 5 m and it makes an angle of 73° with the xaxis.
Question 2:
Let
A→ and B→be the two vectors of magnitude 10 unit each. If they are inclined to the Xaxis at angle 30° and 60° respectively, find the resultant.
Answer:
Angle between
A→ and B→, θ = 60° − 30° = 30°
A→=B→=10 unitsThe magnitude of the resultant vector is given byR=A2+A2+2AAcosθ =102+102+2×10×10×cos 30° =200+200 cos 30° =200+100 =300=17.3 units
Let β be the angle between
R→ and A→.
∴β=tan1 A sin 30°A+A cos 30°⇒β=tan1 10 sin 30°10+10 cos 30°⇒β=tan1 12+3=tan1 13.372⇒β=tan1 0.26795=15°Angle made by the resultant vector with the Xaxis = 15° + 30° = 45°
∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the Xaxis.
Question 3:
Add vectors
A→, B→ and C→each having magnitude of 100 unit and inclined to the Xaxis at angles 45°, 135° and 315° respectively.
Answer:
First, we will find the components of the vector along the xaxis and yaxis. Then we will find the resultant x and ycomponents.
xcomponent of
A→=Acos45°=100 cos 45°=1002 unitxcomponent of
B→ = B→cos 135°=1002xcomponent of
C→=
C→cos315
°= 100 cos 315°
=100 cos 45°=1002Resultant xcomponent
=10021002+1002=1002
Now, ycomponent of
A→=100 sin 45°=1002ycomponent of
B→=100 sin 135°=1002ycomponent of
C→ = 100 sin 315°=1002Resultant ycomponent
=1002+10021002=1002Magnitude of the resultant
=10022+10022
=10000=100Angle made by the resultant vector with the xaxis is given by
tan α=y compx comp =10021002=1⇒ α = tan^{−1} (1) = 45°
∴ The magnitude of the resultant vector is 100 units and it makes an angle of 45° with the xaxis.
Question 4:
Let
a→=4i→+3j→ and b→=3i→+4j→. (a) Find the magnitudes of (a)
a→, (b)
b→, (c)
a→+b→ and(d)
a→b→.
Answer:
Given:
a→=4i→+3j→ and b→=3i→+4j→(a) Magnitude of
a→is given by
a→=42+32=16+9=5(b) Magnitude of
b→is given by
b→=32+42=9+16=5(c)
a→+b→=(4i^+3j^)+(3i^+4j^)=(7i^+7j^)∴ Magnitude of vector
a→+b→is given by
a→+b→=49+49=98=72(d)
a→b→=4i→+3j→3i→+4j→=i→j→∴ Magnitude of vector
a→b→is given by
a→b→=12+12=2
Question 5:
Refer to figure (2E1). Find (a) the magnitude, (b) x and y component and (c) the angle with the Xaxis of the resultant of
OA→, BC→ and DE→.
Figure (2E1)
Answer:
First, let us find the components of the vectors along the x and yaxes. Then we will find the resultant x and ycomponents.
xcomponent of
OA→= 2 cos 30°=3m
xcomponent of
BC→=1.5cos120°
=1.52=7.5m
xcomponent of
DE→= 1cos270°
= 1 × 0 = 0 m
ycomponent of
OA→= 2 sin 30° = 1
ycomponent of
BC→= 1.5 sin 120°
=3×1.52=1.3ycomponent of
DE→= 1 sin 270° = −1
xcomponent of resultant
Rx=30.75+0=0.98 mycomponent of resultant R_{y} = 1 + 1.3 − 1 = 1.3 m
∴Resultant, R=Rx2+Ry2 = 0.982+1.32 =0.96+1.69 =2.65 =1.6 mIf it makes an angle α with the positive xaxis, then
tan α=ycomponentxcomponent =1.30.98=1.332∴ α = tan^{−1} (1.32)
Question 6:
Two vectors have magnitudes 2 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.
Answer:
Let the two vectors be
a→and
b→.
Now,
a→=3 and b→=4(a) If the resultant vector is 1 unit, then
a→2+b→2+2.a→.b→ cos θ=1⇒32+42+2.3.4 cos θ=1Squaring both sides, we get:
25+24 cos θ=1⇒24 cos θ=24⇒cos θ=1⇒θ=180°Hence, the angle between them is 180°.
(b) If the resultant vector is 5 units, then
a→2+b→2+2.a→.b→ cos θ=5⇒32+42+2.3.4 cos θ=5Squaring both sides, we get:
25 + 24 cos θ = 25
⇒ 24 cos θ = 0
⇒ cos θ = 90°
Hence, the angle between them is 90°.
(c) If the resultant vector is 7 units, then
a→2+b→2+2.a→.b→ cos θ=1⇒32+42+2.3.4 cos θ=7Squaring both sides, we get:
25 + 24 cos θ = 49,
⇒ 24 cos θ = 24
⇒ cos θ = 1
⇒ θ = cos^{−1} 1 = 0°
Hence, the angle between them is 0°.
Question 7:
A spy report about a suspected car reads as follows. “The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped”. Find the displacement of the car.
Answer:
The displacement of the car is represented by
AD→.
AD→=2i^+0.5j^+4i^ =6i^+0.5j^Magnitude of
AD→is given by
AD=AE2+DE2 =62+0.52 =36+0.25=6.02 km
Now,
tan θ=DEAE=112
⇒θ=tan1 112Hence, the displacement of the car is 6.02 km along the direction
tan1 112with positive the xaxis.
Question 8:
A carrom board (4 ft × 4 ft square) has the queen at the centre. The queen, hit by the striker moves to the from edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the front edge, (b) from the front edge to the hole and (c) from the centre to the hole.
Answer:
Consider that the queen is initially at point A as shown in the figure.
Let AB be x ft.
So, DE = (2
x) ft
In âˆ†ABC, we have:
tan θ=x2 …(i)
Also, in âˆ†DCE, we have:
tan θ=2x4 …(ii)
From (i) and (ii), we get:
x2=2x4⇒22x=4x⇒42x=4x⇒6x=4⇒x=23ft(a) In âˆ†ABC, we have:
AC=AB2+BC2
=232+22=49+4=409=2310 ft(b) In âˆ†CDE, we have:
DE
=223=623=43 ftCD = 4 ft
∴ CE=CD2+DE2=42+432=4310 ft(c) In âˆ†AGE, we have:
AE=AG2+GE2
=22+22=8+22 ft
Question 9:
A mosquito net over a 7 ft × 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the Xaxis, it width as the Yaxis, and vertically up as the Zaxis, write the components of the displacement vector.
Answer:
Displacement vector of the mosquito,
r→=7i^+4i^+3k^
(a) Magnitude of displacement
=72+42+32
=74 ft(b) The components of the displacement vector are 7 ft, 4 ft and 3 ft along the X, Y and Zaxes, respectively.
Question 10:
Suppose
a→is a vector of magnitude 4.5 units due north. What is the vector (a)
3a→, (b)
4a→?
Answer:
Given:
a→is a vector of magnitude 4.5 units due north.
Case (a):
3a→=3×4.5=13.5∴
3a→is a vector of magnitude 13.5 units due north.
Case (b):
4a→=4×1.5=6 units∴
4a→is a vector of magnitude 6 units due south.
Question 11:
Two vectors have magnitudes 2 m and 3m. The angle between them is 60°. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.
Answer:
Let the two vectors be
a→= 2 m and b→=3 m.
Angle between the vectors, θ = 60°
(a) The scalar product of two vectors is given by
a→.b→=a→.b→ cosθ°.
∴
a→.b→=a→.b→ cos 60°
=2×3×12=3 m2(b) The vector product of two vectors is given by
a→×b→=a→ a→ sinθ°.
∴
a→×b→=a→ a→ sin 60°
=2×3×32=33 m2
Question 12:
Let A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{1} be a regular hexagon. Write the xcomponents of the vectors represented by the six sides taken in order. Use the fact the resultant of these six vectors is zero, to prove that
cos 0 + cos π/3 + cos 2π/3 + cos 3π/3 + cos 4π/3 + cos 5π/3 = 0.
Use the known cosine values to verify the result.
Answer:
According to the polygon law of vector addition, the resultant of these six vectors is zero.
Here, a = b = c = d = e = f (magnitudes), as it is a regular hexagon. A regular polygon has all sides equal to each other.
So,
Rx=A cos 0 + A cos π3+A cos 2π3+A cos 3π3+ A cos 4π3+A cos 5π3=0[As the resultant is zero, the xcomponent of resultant R_{x} is zero]
⇒ cos 0 + cos π3+cos 2π3+cos3π3+cos 4π3+cos 5π5=0
Note: Similarly, it can be proven that
sin 0+sin π3+sin 2π3+sin 3π3+sin 4π3+sin 5π3=0.
Question 13:
Let
a→=2i→+3j→+4k→ and b→=3i→+4j→+5k→. Find the angle between them.
Answer:
We have:
a→=2i→+3j→+4k→b→=3i→+4j→+5k→Using scalar product, we can find the angle between vectors
a→and
b→.
i.e.,
a→.b→=a→b→ cos θSo,
θ=cos1a→.b→a→b→ =cos12×3+3×4+4×522+32+42 32+42+52=cos13829 50=cos1381450∴ The required angle is
cos1381450.
Question 14:
Prove that
A→.A→×B→=0.
Answer:
To prove:
A→.A→×B→=0Proof: Vector product is given by
A→×B→= A→B→ sin n^.
A→B→ sin n^is a vector which is perpendicular to the plane containing
A→ and B→. This implies that it is also perpendicular to
A→. We know that the dot product of two perpendicular vectors is zero.
∴
A→.A→×B→=0Hence, proved.
Question 15:
If
A→=2i→+3j→+4k→ and B→=4i→+3j→+2k→, find
A→×B→.
Answer:
Given:
A→=2i^+3j^+4k^and
B→=4i^+3j^+2k^The vector product of
A→×B→can be obtained as follows:
A→×B→=i^j^k^234432 =i^612j^416+k^612 =6i^+12j^6k^
Question 16:
If
A→, B→, C→are mutually perpendicular, show that
C→×A→×B→=0. Is the converse true?
Answer:
Given:
A→, B→ and C→are mutually perpendicular.
A→×B→is a vector with its direction perpendicular to the plane containing
A→ and B→.
∴ The angle between
C→ and A→×B→is either 0° or 180°.
i.e.,
C→×A→×B→=0However, the converse is not true. For example, if two of the vectors are parallel, then also,
C→×A→×B→=0
So, they need not be mutually perpendicular.
Question 17:
A particle moves on a given straight line with a constant speed ν. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that
OP→×ν→is independent of the position P.
Answer:
The particle moves on the straight line XX’ at a uniform speed ν.
In
∆POQ, we have:
OQ = OP sin θ
OP→×ν=OP ν sin θ n^ =ν OP sin θ n^ =νOQ n^This product is always equal to the perpendicular distance from point O. Also, the direction of this product remains constant.
So, irrespective of the the position of the particle, the magnitude and direction of
OP→×ν→remain constant.
∴OP→ ×ν→is independent of the position P.
Question 18:
The force on a charged particle due to electric and magnetic fields is given by
F→=qE→+qν→×B→. Suppose
E→is along the Xaxis and
B→along the Yaxis. In what direction and with what minimum speed ν should a positively charged particle be sent so that the net force on it is zero?
Answer:
According to the problem, the net electric and magnetic forces on the particle should be zero.
i.e.,
F→=qE→+qν→×B→=0
⇒E=ν→ ×B→So, the direction of
ν→ ×B→should be opposite to the direction of
E→. Hence,
ν→should be along the positive zdirection.
Again, E = νB sin θ
⇒ν=EB sin θFor ν to be minimum,
θ=90° and, thus, νmin=EBSo, the particle must be projected at a minimum speed of
EBalong the zaxis.
Question 19:
Give an example for which
A→·B→=C→·B→ but A→ ≠C→.
Answer:
To prove:
A→·B→=C→·B,→ but A→ ≠C→Suppose that
A→is perpendicular to
B;→ B→is along the west direction.
Also,
B→is perpendicular to
C;→ A→and
C→are along the south and north directions, respectively.
A→is perpendicular to
B→, so there dot or scalar product is zero.
i.e.,
A→·B→=A→B→cosθ=A→B→cos90°=0
B→is perpendicular to
C→, so there dot or scalar product is zero.
i.e.,
C→·B→=C→B→cosθ=C→B→cos90°=0 ∴A→·B→=B→·C, →but A→ ≠C→Hence, proved.
Question 20:
Draw a graph from the following data. Draw tangents at x = 2, 6 and 8. Find the slopes of these tangents. Verify that the curve draw is y = 2x^{2} and the slope of tangent is
tan θ=dydx=4x.
x12345678910y281832507298128162200
Answer:
Note: Students should draw the graph y = 2x^{2} on a graph paper for results.
To find a slope at any point, draw a tangent at the point and extend the line to meet the xaxis. Then find tan θ as shown in the figure.
The above can be checked as follows:
Slope = tan θ=dydx =ddx2x2=4xHere, x = xcoordinate of the point where the slope is to be measured
Question 21:
A curve is represented by y = sin x. If x is changed from
π3toπ3+π100, find approximately the change in y.
Answer:
y = sin x …(i)
Now, consider a small increment âˆ†x in x.
Then y + âˆ†y = sin (x + âˆ†x) …(ii)
Here, âˆ†y is the small change in y.
Subtracting (ii) from (i), we get:
âˆ†y = sin (x + âˆ†x) − sin x
=sin π3+π100sin π3 = 0.0157
Question 22:
The electric current in a charging R−C circuit is given by i = i_{0} e^{−t}^{/RC} where i_{0}, R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) t = 0, (b) t = RC, (c) t = 10 RC.
Answer:
Given: i = i_{0} e^{−t}^{/RC}
∴ Rate of change of current
=didt
=i0 1RC et/RC
=i0RC×et/RCOn applying the conditions given in the questions, we get:
(a) At
t=0, didt=i0RC×e0=i0RC(b) At
t=RC, didt=i0RC×e1=i0RCe(c) At
t=10 RC, dtdi=i0RC×e10=i0RCe10
Question 23:
The electric current in a discharging R−C circuit is given by i = i_{0} e^{−t}^{/RC} where i_{0}, R and C are constant parameters and t is time. Let i_{0} = 2⋅00 A, R = 6⋅00 × 10^{5} Ω and C = 0⋅500 μF. (a) Find the current at t = 0⋅3 s. (b) Find the rate of change of current at at 0⋅3 s. (c) Find approximately the current at t = 0⋅31 s.
Answer:
Electric current in a discharging RC circuit is given by the below equation:
i = i_{0} ⋅ e^{−t}^{/RC} …(i)
Here, i_{0} = 2.00 A
R = 6 × 10^{5} Ω
C = 0.0500 × 10^{−6} F
= 5 × 10^{−7} F
On substituting the values of R, C and i_{0} in equation (i), we get:
i = 2.0 e^{−t}^{/0.3} …(ii)
According to the question, we have:
(a) current at t = 0.3 s
i=2 ×e1=2e A(b) rate of change of current at t = 0.3 s
didt=i0RC·et/RCWhen t = 0.3 s, we have:
didt=20.30·e0.30.3=203e A/s(c) approximate current at t = 0.31 s
i=2e0.30.3 =5.83e A approx.
Page No 30:
Question 24:
Find the area bounded under the curve y = 3x^{2} + 6x + 7 and the Xaxis with the ordinates at x = 5 and x = 10.
Answer:
The given equation of the curve is y = 3x^{2} + 6x + 7.
The area bounded by the curve and the Xaxis with coordinates x_{1} = 5 and x_{2} = 10 is given by
∫x1x2 y dx=∫510 3×2+6x+7dx=3×33+6×22+7×510=1000125+30075+7035=1370235=1135 sq. units
Question 25:
Find the area enclosed by the curve y = sin x and the Xaxis between x = 0 and x = π.
Answer:
The given equation of the curve is y = sin x.
The required area can found by integrating y w.r.t x within the proper limits.
∴Area=∫x1x2 y dx =∫0π sin x dx =cos x0π =cos π –cos 0 =1+1=2 sq. unit
Question 26:
Find the area bounded by the curve y = e^{−x}, the Xaxis and the Yaxis.
Answer:
The given function is y = e^{−x}.
When x = 0, y = e^{−0} = 1
When x increases, the value of y decrease. Also, only when x = ∞, y = 0
So, the required area can be determined by integrating the function from 0 to ∞.
∴Area=∫0∞ex dx =ex0∞ =01=1 sq. unit
Question 27:
A rod of length L is placed along the Xaxis between x = 0 and x = L. The linear density (mass/length) ρ of the rod varies with the distance x from the origin as ρ = a + bx. (a) Find the SI units of a and b. (b) Find the mass of the rod in terms of a, b and L.
Answer:
ρ = mass/length = a + bx
So, the SI unit of ρ is kg/m.
(a)
SI unit of a = kg/m
SI unit of b = kg/m^{2}
(From the principle of homogeneity of dimensions)
(b) Let us consider a small element of length dx at a distance x from the origin as shown in the figure given below:
dm = mass of the element
= ρdx
= (a + dx) dx
∴ Mass of the rod = ∫ dm
=∫0L a+bx dx=ax+bx220L=aL+bL22
Question 28:
The momentum p of a particle changes with time t according to the relation
dpdt=10 N+2 N/st. If the momentum is zero at t = 0, what will the momentum be at t = 10 s?
Answer:
According to the question, we have:
dpdt=10 N+2 N/stMomentum is zero at time, t = 0
Now, dp = [(10 N) + (2 Ns^{−1})t]dt
On integrating the above equation, we get:
p=∫010 dp=∫010 10 dt+∫010 2t dt⇒p=10t+2t22010⇒p=10×10+1000∴ p=100+100=200 kg m/s
Question 29:
The changes in a function y and the independent variable x are related as
dydx=x2. Find y as a function of x.
Answer:
Changes in a function of y and the independent variable x are related as follows:
dydx=x2⇒dy=x2 dxIntegrating of both sides, we get:
∫dy = ∫x^{2} dx
⇒y=x33+c, where c is a constant
∴ y as a function of x is represented by
y=x33+c.
Question 30:
Write the number of significant digits in (a) 1001, (b) 100.1, (c) 100.10, (d) 0.001001.
Answer:
(a) 1001
Number of significant digits = 4
(b) 100.1
Number of significant digits = 4
(c) 100.10
Number of significant digits = 5
(d) 0.001001
Number of significant digits = 4
Question 31:
A metre scale is graduated at every millimetre. How many significant digits will be there in a length measurement with this scale?
Answer:
The metre scale is graduated at every millimetre.
i.e., 1 m = 1000 mm
The minimum number of significant digits may be one (e.g., for measurements like 4 mm and 6 mm) and the maximum number of significant digits may be 4 (e.g., for measurements like 1000 mm). Hence, the number of significant digits may be 1, 2, 3 or 4.
Question 32:
Round the following numbers to 2 significant digits.
(a) 3472, (b) 84.16. (c)2.55 and (d) 28.5
Answer:
(a) In 3472, 7 comes after the digit 4. Its value is greater than 5. So, the next two digits are neglected and 4 is increased by one.
∴ The value becomes 3500.
(b) 84
(c) 2.6
(d) 29
Question 33:
The length and the radius of a cylinder measured with a slide callipers are found to be 4.54 cm and 1.75 cm respectively. Calculate the volume of the cylinder.
Answer:
Length of the cylinder, l = 4.54 cm
Radius of the cylinder, r = 1.75 cm
Volume of the cylinder, V =
πr^{2}l
= (
π) × (4.54) × (1.75)^{2}
The minimum number of significant digits in a particular term is three. Therefore, the result should have three significant digits, while the other digits should be rounded off.
∴ Volume, V =
πr^{2}l
= (3.14) × (1.75) × (1.75) × (4.54)
= 43.6577 cm^{3}
Since the volume is to be rounded off to 3 significant digits, we have:
V = 43.7 cm^{3}
Question 34:
The thickness of a glass plate is measured to be 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the average thickness of the plate from this data.
Answer:
Average thickness=2.17+2.17+2.183 =2.1733 mm∴ Rounding off to three significant digits, the average thickness becomes 2.17 mm.
Question 35:
The length of the string of a simple pendulum is measured with a metre scale to be 90.0 cm. The radius of the bod plus the length of the hook is calculated to be 2.13 cm using measurements with a slide callipers. What is the effective length of the pendulum? (The effective length is defined as the distance between the point of suspension and the centre of the bob.)
Answer:
Consider the figure shown below:
Actual effective length = (90.0 + 2.13) cm
However, in the measurement 90.0 cm, the number of significant digits is only two.
So, the effective length should contain only two significant digits.
i.e., effective length = 90.0 + 2.13 = 92.1 cm.
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity
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