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NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Straight Lines

Edited By Komal Miglani | Updated on May 06, 2025 03:12 PM IST

Have you ever wondered about looking at railway tracks – how they run parallel to each other, or roads forming angles to each other and intersecting at a point? That's the study matter of the chapter on straight lines, where various dimensions such as slopes, types of lines, angles between lines and their various forms of reactions are discussed in detail.

The miscellaneous exercise of Chapter 9 provided in the NCERT acts as a source to practise these concepts in detail. This exercise provided questions of mixed nature from the topics discussed above. This exercise might appear more difficult than the previous one, as the questions provided here are of mixed concepts in nature, but we have made it simple by providing the solutions in a step-by-step manner. Understanding of these concepts given in the Straight Lines strong foundation for more advanced topics in mathematics, such as coordinate geometry, vectors, and calculus. The NCERT solutions provided here are a useful exercise to get conceptual clarity on the topic of straight lines.

This Story also Contains
  1. Class 11 Maths Chapter 1 Straight Lines Miscellaneous Exercise Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 9: Miscellaneous Exercise
  3. Topics covered in Chapter 9 Straight lines Miscellaneous Exercise
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions


Class 11 Maths Chapter 1 Straight Lines Miscellaneous Exercise Solutions - Download PDF

Download PDF


NCERT Solutions Class 11 Maths Chapter 9: Miscellaneous Exercise

Question1: (a) Find the values of k for which the line (k3)x(4k2)y+k27k+6=0 is

Parallel to the x-axis.

Answer:

Given equation of line is
(k3)x(4k2)y+k27k+6=0
and equation of x-axis is y=0
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of y=0 is , m=0
and
Slope of line (k3)x(4k2)y+k27k+6=0 is , m=k34k2
Now,
m=m
k34k2=0
k3=0
k=3
Therefore, the value of k is 3

Question1: (b) Find the values of k for which the line (k3)x(4k2)y+k27k+6=0 is

Parallel to the y-axis.

Answer:

Given the equation of line is
(k3)x(4k2)y+k27k+6=0
and equation of y-axis is x=0
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of y=0 is , m==10
and
Slope of line (k3)x(4k2)y+k27k+6=0 is , m=k34k2
Now,
m=m
k34k2=10
4k2=0
k=±2
Therefore, value of k is ±2

Question1: (c) Find the values of k for which the line (k3)x(4k2)y+k27k+6=0 is Passing through the origin.

Answer:

Given equation of line is
(k3)x(4k2)y+k27k+6=0
It is given that it passes through origin (0,0)
Therefore,
(k3).0(4k2).0+k27k+6=0
k27k+6=0
k26kk+6=0
(k6)(k1)=0
k=6 or 1
Therefore, value of k is 6 or 1


Question 2: Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and 6, respectively.

Answer:

Let the intercepts on x and y-axis are a and b respectively
It is given that
a+b=1  and  a.b=6
a=1b
b.(1b)=6
bb2=6
b2b6=0
b23b+2b6=0
(b+2)(b3)=0
b=2 and 3
Now, when b=2a=3
and when b=3a=2
We know that the intercept form of the line is
xa+yb=1

Case (i) when a = 3 and b = -2
x3+y2=1
2x3y=6

Case (ii) when a = -2 and b = 3
x2+y3=1
3x+2y=6
Therefore, equations of lines are 2x3y=6 and 3x+2y=6

Question 3: What are the points on the y-axis whose distance from the line x3+y4=1 is 4 units.

Answer:

Given the equation of the line is
x3+y4=1
we can rewrite it as
4x+3y=12
Let's take point on y-axis is (0,y)
It is given that the distance of the point (0,y) from line 4x+3y=12 is 4 units
Now,
d=|Ax1+By1+CA2+B2|
In this problem A=4,B=3,C=12,d=4  and  (x1,y1)=(0,y)
4=|4×0+3×y1242+32|=|3y1216+9|=|3y125|

Case (i)

4=3y125
20=3y12
y=323
Therefore, the point is (0,323) -(i)

Case (ii)

4=(3y125)
20=3y+12
y=83
Therefore, the point is (0,83) -(ii)
Therefore, points on the y-axis whose distance from the line x3+y4=1 is 4 units are (0,323) and (0,83)

Question 4: Find perpendicular distance from the origin to the line joining the points (cosθ,sinθ) and (cosϕ,sinϕ)..

Answer:

Equation of line passing through the points (cosθ,sinθ) and (cosϕ,sinϕ) is
(ysinθ)=sinϕsinθcosϕcosθ(xcosθ)
(cosϕcosθ)(ysinθ)=(sinϕsinθ)(xcosθ)
y(cosϕcosθ)sinθ(cosϕcosθ)=x(sinϕsinθ)cosθ(sinϕsinθ)
x(sinϕsinθ)y(cosϕcosθ)=cosθ(sinϕsinθ)sinθ(cosϕcosθ)x(sinϕsinθ)y(cosϕcosθ)=sin(θϕ)
(cosasinbsinacosb=sin(ab))
Now, distance from origin(0,0) is
d=|(sinϕsinθ).0(cosϕcosθ).0sin(θϕ)(sinϕsinθ)2+(cosϕcosθ)2|
d=|sin(θϕ)(sin2ϕ+cos2ϕ)+(sin2θ+cos2θ)2(cosθcosϕ+sinθsinϕ)|
d=|sin(θϕ)1+12cos(θϕ)| (cosacosb+sinasinb=cos(ab)  and  sin2a+cos2a=1)
d=|sin(θϕ)2(1cos(θϕ))|
d=|2sinθϕ2cosθϕ22.(2sin2θϕ2)|
d=|2sinθϕ2cosθϕ22sinθϕ2|
d=|cosθϕ2|

Question 5: Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x7y+5=0 and 3x+y=0.

Answer:

Point of intersection of the lines x7y+5=0 and 3x+y=0
(522,1522)
It is given that this line is parallel to y - axis i.e. x=0 which means their slopes are equal
Slope of x=0 is ,m==10
Let the Slope of line passing through point (522,1522) is m
Then,
m=m=10
Now, equation of line passing through point (522,1522) and with slope 10 is
(y1522)=10(x+522)
x=522
Therefore, equation of line is x=522

Question 6: Find the equation of a line drawn perpendicular to the line x4+y6=1 through the point, where it meets the y-axis.

Answer:

given equation of line is
x4+y6=1
we can rewrite it as
3x+2y=12
Slope of line 3x+2y=12 , m=32
Let the Slope of perpendicular line is m
m=1m=23
Now, the ponit of intersection of 3x+2y=12 and x=0 is (0,6)
Equation of line passing through point (0,6) and with slope 23 is
(y6)=23(x0)
3(y6)=2x
2x3y+18=0
Therefore, equation of line is 2x3y+18=0

Question 7: Find the area of the triangle formed by the lines yx=0,x+y=0 and xk=0.

Answer:

Given equations of lines are
yx=0           (i)
x+y=0           (ii)
xk=0           (iii)
The point if intersection of (i) and (ii) is (0,0)
The point if intersection of (ii) and (iii) is (k,-k)
The point if intersection of (i) and (iii) is (k,k)
Therefore, the vertices of triangle formed by three lines are (0,0),(k,k) and (k,k)
Now, we know that area of triangle whose vertices are (x1,y1),(x2,y2) and (x3,y3) is
A=12|x1(y2y3)+x2(y3y1)+x3(y1y2)|
A=12|0(kk)+k(k0)+k(0+k)|
A=12|k2+k2|
A=12|2k2|
A=k2
Therefore, area of triangle is k2 square units

Question 8: Find the value of p so that the three lines 3x+y2=0,px+2y3=0 and 2xy3=0 may intersect at one point.

Answer:

Point of intersection of lines 3x+y2=0 and 2xy3=0 is (1,1)
Now, (1,1) must satisfy equation px+2y3=0
Therefore,
p(1)+2(1)3=0
p23=0
p=5
Therefore, the value of p is 5

Question 9: If three lines whose equations are y=m1x+c1,y=m2x+c2 and y=m3x+c3 are concurrent, then show that m1(c2c3)+m2(c3c1)+m3(c1c2)=0.

Answer:

Concurrent lines means they all intersect at the same point
Now, given equation of lines are
y=m1x+c1           (i)
y=m2x+c2           (ii)
y=m3x+c3           (iii)
Point of intersection of equation (i) and (ii) (c2c1m1m2,m1c2m2c1m1m2)

Now, lines are concurrent which means point (c2c1m1m2,m1c2m2c1m1m2) also satisfy equation (iii)
Therefore,

m1c2m2c1m1m2=m3.(c2c1m1m2)+c3

m1c2m2c1=m3(c2c1)+c3(m1m2)

m1(c2c3)+m2(c3c1)+m3(c1c2)=0

Hence proved

Question 10: Find the equation of the lines through the point (3,2) which make an angle of 45 with the line x2y=3.

Answer:

Given the equation of the line is
x2y=3
The slope of line x2y=3 , m2=12
Let the slope of the other line is, m1=m
Now, it is given that both the lines make an angle 45 with each other
Therefore,
tanθ=|m2m11+m1m2|
tan45=|12m1+m2|
1=|12m2+m|
Now,

Case (i)
1=12m2+m
2+m=12m
m=13
Equation of line passing through the point (3,2) and with slope 13
(y2)=13(x3)
3(y2)=1(x3)
x+3y=9                (i)

Case (ii)
1=(12m2+m)
2+m=(12m)
m=3
Equation of line passing through the point (3,2) and with slope 3 is
(y2)=3(x3)
3xy=7               (ii)

Therefore, equations of lines are 3xy=7 and x+3y=9

Question 11: Find the equation of the line passing through the point of intersection of the lines 4x+7y3=0 and 2x3y+1=0 that has equal intercepts on the axes.

Answer:

Point of intersection of the lines 4x+7y3=0 and 2x3y+1=0 is (113,513)
We know that the intercept form of the line is
xa+yb=1
It is given that line make equal intercepts on x and y axis
Therefore,
a = b
Now, the equation reduces to
x+y=a -(i)
It passes through point (113,513)
Therefore,
a=113+513=613
Put the value of a in equation (i)
we will get
13x+13y=6
Therefore, equation of line is 13x+13y=6

Question 12: Show that the equation of the line passing through the origin and making an angle θ with the line y=mx+c is yx=m±tanθ1mtanθ.

Answer:

Slope of line y=mx+c is m
Let the slope of other line is m'
It is given that both the line makes an angle θ with each other
Therefore,
tanθ=|m2m11+m1m2|
tanθ=|mm1+mm|
(1+mm)tanθ=(mm)
tanθ+m(mtanθ+1)=m
m=m±tanθ1mtanθ
Now, equation of line passing through origin (0,0) and with slope m±tanθ1mtanθ is
(y0)=m±tanθ1mtanθ(x0)
yx=m±tanθ1mtanθ
Hence proved

Question 13: In what ratio, the line joining (1,1) and (5,7) is divided by the line x+y=4 ?

Answer:

Equation of line joining (1,1) and (5,7) is
(y1)=715+1(x+1)
(y1)=66(x+1)
(y1)=1(x+1)
xy+2=0
Now, point of intersection of lines x+y=4 and xy+2=0 is (1,3)
Now, let's suppose point (1,3)divides the line segment joining (1,1) and (5,7) in 1:k
Then,
(1,3)=(k(1)+1(5)k+1,k(1)+1(7)k+1)
1=k+5k+1  and  3=k+7k+1
k=2
Therefore, the line joining (1,1) and (5,7) is divided by the line x+y=4 in ratio 1:2

Question 14: Find the distance of the line 4x+7y+5=0 from the point (1,2) along the line 2xy=0.

Answer:

22229
point (1,2) lies on line 2xy=0
Now, point of intersection of lines 2xy=0 and 4x+7y+5=0 is (518,59)
Now, we know that the distance between two point is given by
d=|(x2x1)2+(y2y1)2|
d=|(1+518)2+(2+59)2|
d=|(2318)2+(239)2|
d=|529324+52981|
d=|529+2116324|=|2645324|=23518
Therefore, the distance of the line 4x+7y+5=0 from the point (1,2) along the line 2xy=0 is 23518 units

Question 15: Find the direction in which a straight line must be drawn through the point (1,2) so that its point of intersection with the line x+y=4 may be at a distance of 3 units from this point.

Answer:

Let (x1,y1) be the point of intersection
it lies on line x+y=4
Therefore,
x1+y1=4x1=4y1           (i)
Distance of point (x1,y1) from (1,2) is 3
Therefore,
3=|(x1+1)2+(y12)2|
Square both the sides and put value from equation (i)
9=(5y1)2+(y12)29=y12+2510y1+y12+44y12y1214y1+20=0y127y1+10=0y125y12y1+10=0(y12)(y15)=0y1=2 or y1=5
When y1=2x1=2 point is (2,2)
and
When y1=5x1=1 point is (1,5)
Now, slope of line joining point (2,2) and (1,2) is
m=2212=0
Therefore, line is parallel to x-axis -(i)

or
slope of line joining point (1,5) and (1,2)
m=521+2=
Therefore, line is parallel to y-axis -(ii)

Therefore, line is parallel to x -axis or parallel to y-axis

Question 16: The hypotenuse of a right angled triangle has its ends at the points (1,3) and (4,1) Find an equation of the legs (perpendicular sides) of the triangle.

Answer:

22234
Slope of line OA and OB are negative times inverse of each other
Slope of line OA is , m=3y1x(3y)=m(1x)
Slope of line OB is , 1m=1y4x(x+4)=m(1y)
Now,
for a given value of m we get these equations
If m=
1x=0    and     1y=0
x=1    and     y=1

Question 17: Find the image of the point (3,8) with respect to the line x+3y=7 assuming the line to be a plane mirror.

Answer:

Ch. 10
Let point (a,b) is the image of point (3,8) w.r.t. to line x+3y=7
line x+3y=7 is perpendicular bisector of line joining points (3,8) and (a,b)
Slope of line x+3y=7 , m=13
Slope of line joining points (3,8) and (a,b) is , m=8b3a
Now,
m=1m              (lines are perpendicular)
8b3a=3
8b=93a
3ab=1           (i)
Point of intersection is the midpoint of line joining points (3,8) and (a,b)
Therefore,
Point of intersection is (3+a2,b+82)
Point (3+a2,b+82) also satisfy the line x+3y=7
Therefore,
3+a2+3.b+82=7
a+3b=13             (ii)
On solving equation (i) and (ii) we will get
(a,b)=(1,4)
Therefore, the image of the point (3,8) with respect to the line x+3y=7 is (1,4)

Question 18: If the lines y=3x+1 and 2y=x+3 are equally inclined to the line y=mx+4 , find the value of m.

Answer:

Given equation of lines are
y=3x+1          (i)
2y=x+3          (ii)
y=mx+4          (iii)
Now, it is given that line (i) and (ii) are equally inclined to the line (iii)
Slope of line y=3x+1 is , m1=3
Slope of line 2y=x+3 is , m2=12
Slope of line y=mx+4 is , m3=m
Now, we know that
tanθ=|m1m21+m1m2|
Now,
tanθ1=|3m1+3m| and tanθ2=|12m1+m2|

It is given that tanθ1=tanθ2
Therefore,
|3m1+3m|=|12m2+m|
3m1+3m=±(12m2+m)
Now, if 3m1+3m=(12m2+m)
Then,
(2+m)(3m)=(12m)(1+3m)
6+mm2=1+m6m2
5m2=5
m=1
Which is not possible
Now, if 3m1+3m=(12m2+m)
Then,
(2+m)(3m)=(12m)(1+3m)
6+mm2=1m+6m2
7m22m7=0
m=(2)±(2)24×7×(7)2×7=2±20014=1±527

Therefore, the value of m is 1±527

Question 19: If the sum of the perpendicular distances of a variable point P(x,y) from the lines x+y5=0 and 3x2y+7=0 is always 10. Show that P must move on a line.

Answer:

Given the equation of line are
x+y5=0              (i)
3x2y+7=0           (ii)
Now, perpendicular distances of a variable point P(x,y) from the lines are

d1=|1.x+1.y512+12| d2=|3.x2.y+732+22|
d1=|x+y52| d2=|3x2y+713|
Now, it is given that
d1+d2=10
Therefore,
x+y52+3x2y+713=10
(assuming x+y5>0 and 3x2y+7>0)
(x+y5)13+(3x2y+7)2=1026

x(13+32)+y(1322)=1026+51372

Which is the equation of the line
Hence proved

Question 20: Find equation of the line which is equidistant from parallel lines 9x+6y7=0 and 3x+2y+6=0.

Answer:

Let's take the point p(a,b) which is equidistance from parallel lines 9x+6y7=0 and 3x+2y+6=0
Therefore,
d1=|9.a+6.b792+62| d2=|3.a+2.b+632+22|
d1=|9a+6b7117| d2=|3a+2b+613|
It is that d1=d2
Therefore,
|9a+6b7313|=|3a+2b+613|
(9a+6b7)=±3(3a+2b+6)
Now, case (i)
(9a+6b7)=3(3a+2b+6)
25=0
Therefore, this case is not possible

Case (ii)
(9a+6b7)=3(3a+2b+6)
18a+12b+11=0

Therefore, the required equation of the line is 18a+12b+11=0

Question 21: A ray of light passing through the point (1,2) reflects on the x-axis at point A and the reflected ray passes through the point (5,3). Find the coordinates of A.

Answer:

22246
From the figure above we can say that
The slope of line AC (m)=tanθ
Therefore,
tanθ=305a=35a          (i)
Similarly,
The slope of line AB (m)=tan(180θ)
Therefore,
tan(180θ)=201a
tanθ=21a
tanθ=2a1             (ii)
Now, from equation (i) and (ii) we will get
35a=2a1
3(a1)=2(5a)
3a3=102a
5a=13
a=135
Therefore, the coordinates of A. is (135,0)

Question 22: Prove that the product of the lengths of the perpendiculars drawn from the points (a2b2,0) and (a2b2,0) to the line xacosθ+ybsinθ=1 is b2.

Answer:

Given equation id line is
xacosθ+ybsinθ=1
We can rewrite it as
xbcosθ+yasinθ=ab
Now, the distance of the line xbcosθ+yasinθ=ab from the point (a2b2,0) is given by
d1=|bcosθ.a2b2+asinθ.0ab(bcosθ)2+(asinθ)2|=|bcosθ.a2b2ab(bcosθ)2+(asinθ)2|
Similarly,
The distance of the line xbcosθ+yasinθ=ab from the point (a2b2,0) is given by
d2=|bcosθ.(a2b2)+asinθ.0ab(bcosθ)2+(asinθ)2|=|(bcosθ.a2b2+ab)(bcosθ)2+(asinθ)2|
d1.d2=|bcosθ.a2b2ab(bcosθ)2+(asinθ)2|.×|(bcosθ.a2b2+ab)(bcosθ)2+(asinθ)2|
=|((bcosθ.a2b2)2(ab)2)(bcosθ)2+(asinθ)2|
=|b2cos2θ.(a2b2)+a2b2)(bcosθ)2+(asinθ)2|
=|a2b2cos2θ+b4cos2θ+a2b2)b2cos2θ+a2sin2θ|
=|b2(a2cos2θb2cos2θa2)b2cos2θ+a2sin2θ|
=|b2(a2cos2θb2cos2θa2(sin2θ+cos2θ))b2cos2θ+a2sin2θ|    (sin2a+cos2a=1)
=|b2(a2cos2θb2cos2θa2sin2θa2cos2θ)b2cos2θ+a2sin2θ|
=|+b2(b2cos2θ+a2sin2θ)b2cos2θ+a2sin2θ|
=b2
Hence proved

Question 23: A person standing at the junction (crossing) of two straight paths represented by the equations 2x3y+4=0 and 3x+4y5=0 wants to reach the path whose equation is 6x7y+8=0 in the least time. Find equation of the path that he should follow.

Answer:

point of intersection of lines 2x3y+4=0 and 3x+4y5=0 (junction) is (117,2217)
Now, person reaches to path 6x7y+8=0 in least time when it follow the path perpendicular to it
Now,
Slope of line 6x7y+8=0 is , m=67
let the slope of line perpendicular to it is , m
Then,
m=1m=76
Now, equation of line passing through point (117,2217) and with slope 76 is
(y2217)=76(x(117))
6(17y22)=7(17x+1)
102y132=119x7
119x+102y=125

Therefore, the required equation of line is 119x+102y=125

Also read

Background wave

Topics covered in Chapter 9 Straight lines Miscellaneous Exercise

  • Slope of a Line: The slope tells us how steep a line is, showing how much it goes up or down as we move from left to right.

  • Various Forms of the Equation of a Line: The equation of a line can be written differently depending on the information given, like its slope, a point on it, or its x and y values.

  • General Equation of a Line: This is a standard way to write any straight line using constants with the x and y terms.

  • Distance of a Point From a Line: This tells us how far a point is from a line by measuring the shortest path, which is a straight line drawn at a right angle.

  • Area of the Triangle Formed by Lines: When three lines intersect and form a triangle, we can find how much space it covers by using the points where the lines cross.

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Frequently Asked Questions (FAQs)

1. Write the equation of line passing through (5,4) and parallel to y-axis ?

Equation of line parallel to y-axis  =>  x = c

Line passing through the point (5,4)  =>  5 = c

Equation of line   =>  x = 5

2. Write the equation of line passing through (5,4) and parallel to x-axis ?

Equation of line parallel to x-axis  =>  y = c

Line passing through the point (5,4)  =>  4 = c

Equation of line   =>  y = 4

3. Write the equation of line passing through (1,0) and make angle of 45 degree with positive x-axis ?

Line make angle of 45 degree with positive x-axis  => slope of line = 1

Equation of line with slope 1 => y = x + c

Line pass through the (1,0)  =>  0 = 1 + c  =>  c=-1

Equation of line =>  y = x -1

4. Find the slope of line 4y = 3x - 6 ?

Given line 4y = 3x - 6

y = 3x/4 - 6/4

Compare with y = mx + c

Slope (m)  = 3/4

5. Do I need to buy NCERT solutions book for Class 11 Maths chapter 9 straight line ?

No, you don't need to buy the NCERT solutions book for Class 11 Maths chapter 9, straight line. Here you will get NCERT Solutions for Class 11 Maths Chapter 9.

6. Do I need to buy NCERT exemplar solutions book for Class 11 Maths chapter 9 straight line ?

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Option 1)

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Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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