False, Defense: L.H.S \[=\text{ }\left( tan\text{ }\theta +2 \right)\text{ }\left( 2\text{ }tan\text{ }\theta +1 \right)\] \[=\text{ }2\text{ }tan2\text{ }\theta \text{ }+\text{ }tan\text{ }\theta...
True: Reason: As indicated by the inquiry, \[cos\text{ }A+cos2\text{ }A\text{ }=\text{ }1\] i.e., \[cos\text{ }A\text{ }=\text{ }1-cos2\text{ }A\] Since, \[sin2\text{ }\theta +cos2\text{ }\theta...
Solution: True, reason:
False, Defense: We realize that, sin θ increments when 0° ≤ θ ≤ 90° cos θ diminishes when 0° ≤ θ ≤ 90° What's more, (sin 80°-cos 80°) = (expanding esteem diminishing worth) = a positive worth....
False, Reason: Since, \[\left( a2-b2 \right)\text{ }=\text{ }\left( a+b \right)\left( a-b \right)\] \[cos2\text{ }23{}^\circ \text{ }\text{ }sin2\text{ }67{}^\circ \text{ }=\left( cos\text{...
1. tan 47o/cot 43 ° = 1 Valid Legitimization: Since, tan (90° - θ) = bed θ
(A) 1/√3 (B) √3 (C) 1 (D) 0 (C) 1 As indicated by the inquiry, \[cos\text{ }9\propto =\text{ }sin\propto and\text{ }9\propto <90{}^\circ \] for example 9α is an intense point We realize that,...
(A) 0 (B) 1 (C) 2 (D) ½ (B) 1 tan 1°. tan 2°.tan 3° … tan 89° =tan1°.tan 2°.tan 3°… tan 43°.tan 44°.tan 45°.tan 46°.tan 47°… tan 87°.tan 88°.tan 89° Since, tan 45° = 1, = tan1°.tan 2°.tan 3°… tan...
(A) cos β (B) cos 2β (C) sin α (D) sin 2α (B) cos 2β As indicated by the inquiry, \[cos\left( \alpha +\beta \right)\text{ }=\text{ }0\] Since, cos 90° = 0 We can compose, \[cos\left( \alpha...
(A) b/√(b2– a2) (B) b/a (C) √(b2-a2)/b (D) a/√(b2-a2) (C) √(b2 – a2)/b As indicated by the inquiry, \[sin\text{ }\theta \text{ }=a/b\] We know, \[sin2\text{ }\theta \text{ }+cos2\text{ }\theta...
(A) – 1 (B) 0 (C) 1 (D) 3 2 (B) 0 As indicated by the inquiry, We need to discover the worth of the situation, \[cosec\left( 75{}^\circ +\theta \right)\text{ }\text{ }sec\left( 15{}^\circ...
(A) √3 (B) 1/√3 (C) √3/2 (D) 1 (A) √3 As indicated by the inquiry, \[Sin\text{ }A\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\ldots \text{ }\left( 1 \right)\] We realize that, NCERT...
(A) 3/5 (B) ¾ (C) 4/3 (D) 5/3 (B) 3/4 As indicated by the inquiry, \[cos\text{ }A\text{ }=\text{ }4/5\text{ }\ldots \text{ }\left( 1 \right)\] We know, \[tan\text{ }A\text{ }=\text{ }sinA/cosA\] To...
Allow Zeba's to age = x As indicated by the inquiry, \[\left( x-5 \right){}^\text{2}=11+5x\] \[x{}^\text{2}+25-10x=11+5x\] \[x{}^\text{2}-15x+14=0\] \[x{}^\text{2}-14x-x+14=0\] \[x\left( x-14...
Let unique speed of train = x km/h We know, Time = distance/speed As indicated by the inquiry, we have, Time taken via train = 360/x hour What's more, Time taken via train its speed increment 5 km/h...
Let the normal number = 'x'. As per the inquiry, We get the condition, \[x{}^\text{2}\text{ }\text{ }84\text{ }=\text{ }3\left( x+8 \right)\] \[x{}^\text{2}\text{ }\text{ }84\text{ }=\text{...
(½)x2– √11x + 1 = 0 using the formulae: we get,
x2 + 2 √2x – 6 = 0x2 – 3 √5x + 10 = 0 using formulae: 5. we get, 6.
–3x2 + 5x + 12 = 0–x2 + 7x – 10 = 0 using the formulae: 3. we get, 4.
2 x2 – 3x – 5 = 05x2 + 13x + 8 = 0by using ths formulae below: 1. we get, 2.
Given, $AP=10cm$ $\angle APB={{60}^{\circ }}$ According to the figure We know that, A line drawn from centre to point from where external tangents are drawn, bisects the angle made by tangents at...
No, a quadratic condition with essential coefficients might possibly have indispensable roots. Support Think about the accompanying condition, \[8x2\text{ }\text{ }2x\text{ }\text{ }1\text{ }=\text{...
If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.If the coefficient of x2 and the constant term have the same...
Every quadratic equation has at least two roots.Every quadratic equations has at most two roots. (iii) False. For instance, a quadratic condition \[x2\text{ }\text{ }4x\text{ }+\text{ }4\text{...
Every quadratic equation has exactly one root.Every quadratic equation has at least one real root. (I) False. For instance, a quadratic condition \[x2\text{ }\text{ }9\text{ }=\text{ }0\] has two...
(x – 1) (x + 2) + 2 = 0(x + 1) (x – 2) + x = 0 (ix) The condition (x – 1) (x + 2) + 2 = 0 has two genuine and unmistakable roots. Working on the above condition, \[x2\text{ }\text{ }x\text{ }+\text{...
√2 x2 –(3/√2)x + 1/√2 = 0x (1 – x) – 2 = 0 (vii) The condition √2x2 – 3x/√2 + ½ = 0 has two genuine and unmistakable roots. \[\begin{array}{*{35}{l}} D\text{ }=\text{ }b2\text{ }\text{...
(x + 4)2 – 8x = 0(x – √2)2 – 2(x + 1) = 0 (v) The condition (x + 4)2 – 8x = 0 has no genuine roots. Working on the above condition, \[x2\text{ }+\text{ }8x\text{ }+\text{...
2x2 – 6x + 9/2 = 03x2 – 4x + 1 = 0 (iii) The condition 2x2 – 6x + (9/2) = 0 has genuine and equivalent roots. \[D\text{ }=\text{ }b2\text{ }\text{ }4ac\] \[=\text{ }\left( -\text{ }6 \right)2\text{...
x2 – 3x + 4 = 02x2 + x – 1 = 0 (I) The condition x2 – 3x + 4 = 0 has no genuine roots. \[D\text{ }=\text{ }b2\text{ }\text{ }4ac\] \[=\text{ }\left( -\text{ }3 \right)2\text{ }\text{...
(A) 2x2 – 3x + 6 = 0 (B) –x2 + 3x – 3 = 0 (C) √2x2 – 3/√2x+1=0 (D) 3x2 – 3x + 3 = 0 (B) – \[x2\text{ }+\text{ }3x\text{ }\text{ }3\text{ }=\text{ }0\] The amount of the foundations of a...
Given, $\angle TRQ={{30}^{\circ }}$ . At point R, OR ⊥ RQ. So, $\angle ORQ={{90}^{\circ }}$ $\Rightarrow \angle TRQ+\angle ORT={{90}^{\circ }}$ $\Rightarrow \angle ORT={{90}^{\circ...
(A) 2 (B) – 2 (C) ¼ (D) ½ (A) 2 In case ½ is a foundation of the situation \[x2\text{ }+\text{ }kx\text{ }\text{ }5/4\text{ }=\text{ }0\] then, at that point, subbing the worth of ½ instead of x...
(A) x2 – 4x + 5 = 0 (B) x2 + 3x – 12 = 0 (C) 2x2 – 7x + 6 = 0 (D) 3x2 – 6x – 2 = 0 (C) \[2x2\text{ }\text{ }7x\text{ }+\text{ }6\text{ }=\text{ }0\] Assuming 2 is a...
(A) 2(x – 1)2 = 4x2 – 2x + 1 (B) 2x – x2 = x2 + 5 (C) (√2x + √3)2 + x2 = 3x2 − 5x (D) (x2 + 2x)2 = x4 + 3 + 4x3 (D) \[\left( x2\text{ }+\text{ }2x...
We are considering a circle with centre ‘O’ with two parallel tangents through A & B at ends of diameter. M intersect the parallel tangents at P and Q Then, required to prove: $\angle...
Since AB and AC are the tangents from the same point A ∴AB=AC=5cm Similarly, BP=PX and XQ=QC Perimeter of \[\Delta APQ=AP+AQ+PQ\] \[=AP+AQ+(PX+XQ)\] \[=(AP+PX)+(AQ+XQ)\] \[=(AP+BP)+(AQ+QC)\]...
Solution: Let’s consider the original number of people as ‘a’. Given that, $Rs.9000$ were divided equally among a certain number of persons. Had there been $20$ more persons, each would have got...
Solution: Let the original price of the toy be ‘x’. Given that, when the list price of a toy is reduced by $Rs.2$, the person can buy $2$ toys more for $Rs.360$. The number of toys he can buy at the...
Solution: Let the cost price be assumed as Rs x. Given, the dealer sells an article for $Rs.24$ and gains as much percent as the cost price of the article. It’s given that he gains as much as the...
Solution: Let the number of students who planned the picnic be ‘x’. And given, budget for the food was $Rs.480$ So, cost of food for each member $= 480/x$ Also given, eight of these failed to go and...
Solution: Let’s assume the length of the cloth to be ‘a’ meters. Given, piece of cloth costs $Rs.35$ and if the piece were $4m$ longer and each metre costs $Rs.1$ less, the cost remains unchanged....
Provided in question: Chord PQ is parallel to tangent at R.To prove: R bisects the arc PRQ. Proof: Since PQ || tangent at R. $\angle 1=\angle 2$ [alternate interior angles]$\angle 1=\angle 3$...
Solution: Given, ${{a}_{{{n}_{{}}}}}=-4n+15$ Now putting $n = 1, 2, 3, 4$ we get, ${{a}_{1}}=-4[1]+15=-4+15=11$ ${{a}_{2}}=-4[2]+15=-8+15=7$ ${{a}_{3}}=-4[3]+15=-12+15=3$ ...
Suppose C1 and C2 are two circles with the same center O. And AC is a chord touching C1 at the point D let’s join OD.So, $OD\bot AC$$AD=DC=4cm$ [perpendicular line OD...
Solution: Given, ${{a}_{n}}=3{{n}_{2}}-5$ Now putting $n = 1, 2,3,4$ we get,...
Let’s Consider a quadrilateral ABCD touching circle with the centre O at points E, F, G and H as we can see in figure. We know that, In a circle with two points outside of it, the tangents drawn...
(A) x2 + 2x + 1 = (4 – x)2 + 3 (B) –2x2 = (5 – x)(2x-(2/5)) (C) (k + 1) x2 + (3/2) x = 7, where k = –1 (D) x3 – x2 = (x – 1)3 (D) \[x3\text{ }\text{ }x2\text{ }=\text{ }\left( x\text{ }\text{ }1...
Solution: Given, $an = 5n – 7$ Now putting $n = 1, 2, 3, 4$ we get, ${{a}_{1}}=5[1]=5-7=-2$$$ ${{a}_{2=}}5[2]-7=10-7=3$ ${{a}_{3}}=5[3]-7=15-7=8$ ${{a}_{4}}=5[4]-7=20-7=13$ We can see that,...
hour less for a journey of 
if its speed is increased by 
from its usual speed of the plane. Find its usual speed.Solution: ...
Given, OP = $26cm$ PT = tangent length = $10cm$ To find: radius = OT =$?$ We know that, Radius and tangent are perpendicular at the point of contact, $\angle OTP={{90}^{\circ }}$ $$ So,...
Solution: According to Euclid’s division Lemma, Let n be the positive integer And b equals to 3 Where q is the quotient and r is the remainder $n\text{ }=3q+r$, 0<r<3 implies remainders may be...
Solution: Where q is an integer and r = 0, 1, 2, 3, 4, 5, then 6q + r is a positive integer. The positive integers are then of the form: 6q, 6q+1, 6q+2, 6q+3, 6q+4, and 6q+5. Taking cube on L.H.S...
Solution: Let x and y be 2k + 1 and 2p + 1 the two odd positive numbers, respectively Therefore, ${{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }{{\left( 2k+\text{ }1 \right)}^{2}}~+{{\left( 2p\text{ }+1...
Solution: Any odd positive integer n can be written in the form 4q + 1 or 4q + 3 as we know. Now, according to the question, For $n~=4q~+1$, Therefore ${{n}^{2}}~-1\text{ }={{\left( 4q~+\text{ }1...
Solution: Let b=4 and a be any odd integer. According to Euclid’s algorithm, For some integer $m\ge ~0,\text{ }a=4m+r$ And r = 0,1,2,3 as 0 ≤ r < 4. As a result we have, a = 4m or 4m +...
Solution: Let a be the positive integer. According to Euclid’s division algorithm, $a=6q+r$, where the value of 0 ≤ r < 6 ${{a}^{2}}~={{\left( 6q+r \right)}^{2}}~=36{{q}^{2}}~+\text{...
Solution: Let ‘a’ be the positive integer According to Euclid’s division lemma, $a=bm+r$ The value of b is 5 according to the question, $a=5m+r$ So, r= 0, 1, 2, 3, 4 For r = 0,$a=5m$. For r =...
Solution: Let any positive integer be ‘a’ and the value b is 4. According to Euclid Division Lemma, $a=bq+r$, where, 0 ≤ r < b $a=3q+r$, where, 0 ≤ r < 4 The possible values of ‘r’ according...
Solution: According to Euclid’s division lemma, $~a=bq+r$ According to the question, For b = 4. $a=4k+r$, 0 < r < 4 For r = 0, we obtain, $a=4k$...
\[\begin{array}{*{35}{l}} 2x2\text{ }+\left( 7/2 \right)x\text{ }+3/4 \\ ~ \\ \end{array}\] The condition can likewise be composed as, \[8x2+14x+3\] Parting the...
\[t3\text{ }\text{ }2t2\text{ }\text{ }15t\] Taking t normal, we get, \[t\text{ }\left( \text{ }t2\text{ }-\text{ }2t\text{ }-\text{ }15 \right)\] Parting the center term of the situation t2 - 2t -...
Answer- Expression for power of lens (P) = 1/f We are given that P = 1.5D Therefore, f = 1/1.5 f = 10/15 So, f = 0.66 m
\[5t2\text{ }+\text{ }12t\text{ }+\text{ }7\] Parting the center term, we get, \[5t2\text{ }+5t\text{ }+\text{ }7t\text{ }+\text{ }7\] Taking the normal factors out, we get, \[5t\text{ }\left( t+1...
Answer- According to the question, Power of the lens (P) is P = -2D And the expression for the power of the lens is P = 1/f So, we get => f = -1/2 f = -0.5 m As it has a negative focal length....
\[3x2\text{ }+\text{ }4x\text{ }\text{ }4\] Parting the center term, we get, \[3x2\text{ }+\text{ }6x\text{ }\text{ }2x\text{ }\text{ }4\] Taking the normal factors out, we get, \[3x\left( x+2...
1. 4x2 – 3x – 1 \[4x2\text{ }\text{ }3x\text{ }\text{ }1\] Parting the center term, we get, \[4x2-4x+1x-1\] Taking the normal factors out, we get, \[4x\left( x-1 \right)\text{ }+1\left(...
Answer- According to the question, object distance (u) is – 20 cm And object height (h) is 5 cm Radius of curvature (R) is 30 cm And we know that radius of curvature = 2 × Focal length Or, R = 2f We...
Answer- A virtual and erect image generated by a flat mirror is denoted by a positive sign. The size of the image is equal to the size of the object because the magnification is 1.
Answer- According to the focal length of convex mirror, (f) is +15 cm And the object distance (u) is – 10 cm Using the mirror formula, 1/f = 1/v - 1/u Or, 1/v = 1/f + 1/u = 1/15 - 1/(-10) v = 5/30 =...
Answer- According to the question, focal length of concave lens (OF1) is f = – 15 cm Image distance (v) is – 10 cm Using the lens formula - 1/f = 1/v - 1/u Or, 1/v = 1/f + 1/u = - 1/(10) - 1/(-15)...
Given, √2 is one of the zero of the cubic polynomial. Then, at that point, (x-√2) is one of the factor of the given polynomial \[p\left( x \right)\text{ }=\text{ }6x{}^\text{3}+\surd...
Considering that \[a,\text{ }a+b,\text{ }a+2b\] are foundations of given polynomial \[x{}^\text{3}-6x{}^\text{2}+3x+10\] Amount of the roots ⇒ \[a+2b+a+a+b\] = - coefficient of x²/coefficient of x³...
Answer- According to the question, height of the Object h0 is 5 cm And the distance of the object from converging lens (u) is -25 cm Also the Focal length of a converging lens is f = 10 cm...
(iii) -2√3, -9 (iv) (-3/(2√5)), -½ (iii) Sum of the zeroes = – 2√3 Result of the zeroes = – 9 P(x) = x2 – (amount of the zeroes) + (result of the zeroes) Then, at that point, \[P\left( x...
(i) (–8/3), 4/3 (ii) 21/8, 5/16 (I) Sum of the zeroes = – 8/3 Result of the zeroes = 4/3 P(x) = x2 – (amount of the zeroes) + (result of the zeroes) Then, at that point, \[P\left( x \right)=\text{...
(v) A Quadratic Equation will have equivalent roots on the off chance that it fulfills the condition: \[b{}^\text{2}\text{ }\text{ }4ac\text{ }=\text{ }0\] Given condition is \[x{}^\text{2}\text{...
(iii) If on division of a polynomial p (x) by a polynomial g (x), the remainder is zero, what is the connection between the levels of p (x) and g (x)? We realize that, \[p\left( x \right)=\text{...
Answer- Yes, as indicated in the illustration, it will provide a comprehensive image of the object. The image of a distant object, such as a tree. on a screen when the lower part of the lens is...
(a) Headlights of a car (b) Side/rear-view mirror of a vehicle (c) Solar furnace Support your answer with reason. Answer- (a) Concave Because when the light source is placed at their major focus,...
(i) Can x2 – 1 be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5? (ii) What will the quotient and remainder be on division of ax2 + bx + c by px3 + qx2 + rx + s, p ≠...
Answer- The object's distance from the mirror's pole ranges from 0 to 15 cm. The image's nature is virtual, upright, and larger than the thing.
(a) A convex lens of focal length 50 cm (b) A concave lens of focal length 50 cm (c) A convex lens of focal length 5 cm (d) A concave lens of focal length 5 cm Answer – The correct option is (c)....
(A) (–c/a) (B) c/a (C) 0 (D) (–b/a) (B) (c/a) Clarification: As indicated by the inquiry, We have the polynomial, \[ax3\text{ }+\text{ }bx2\text{ }+\text{ }cx\text{ }+\text{ }d\] We realize that,...
(A) 1 (B) 2 (C) 3 (D) more than 3 (D) more than 3 Clarification: As per the inquiry, The zeroes of the polynomials = - 2 and 5 We realize that the polynomial is of the structure, \[p\left( x...
(a) both concave (b) both convex (c) the mirror is concave and the lens is convex (d) the mirror is convex, but the lens is concave Answer – (a) Both concave The focal length of a concave mirror and...
(A) a = –7, b = –1 (B) a = 5, b = –1 (C) a = 2, b = – 6 (D) a = 0, b = – 6 (D) a = 2, b = – 6 Clarification: As per the inquiry, \[x{}^\text{2}\text{ }+\text{ }\left( a+1 \right)x\text{ }+\text{...
(a) At the principal focus of the lens (b) At twice the focal length (c) At infinity (d) Between the optical centre of the lens and its principal focus. Answer – The correct option is (b). The...
(a) Between the principal focus and the centre of curvature (b) At the centre of curvature (c) Beyond the centre of curvature (d) Between the pole of the mirror and its principal focus. Answer- The...
(a) Water (b) Glass (c) Plastic (d) Clay Answer – The correct option is (d). Clay cannot be used to create a lens because light rays cannot flow through it.
(A) x2 – x + 12 (B) x2 + x + 12 (C) (x2/2)-(x/2)-6 (D) 2x2 + 2x –24 (C) \[\left( x2/2 \right)-\text{ }\left( x/2 \right)-\text{ }6\] Clarification: Amount of zeroes, \[\alpha +\text{ }\beta =\text{...
Answer- According to the question, focal length of concave lens (f) is 2 m Power of lens (P) = 1/f Power = 1/ (-2) Therefore, Power = -0.5D
Answer- Because the image is actual and the same size, it should be positioned at 2F. The image of the needle is assumed to be formed at a distance of 50 cm from the convex lens. As a result, the...
(A) 4/3 (B) -4/3 2/3 (D) -2/3 (A) 4/3 Clarification: As indicated by the inquiry, - 3 is one of the zeros of quadratic polynomial \[\left( k-1 \right)x2+kx+1\] Subbing - 3 in the given polynomial,...
km was 
more than the time taken in the return journey. If he returned at the speed of 
more than the speed of going, what was the speed per hour in each direction?Solution: Let the ongoing speed of person be x km/hr, Then, the returning speed of the person is $= (x + 10) km/hr$ (from the question) Using, speed = distance/ time Time taken by the person in...
Answer- The symbol D stands for dioptre, which is the SI unit of lens power. A dioptre is the power of a lens with a focal length of one metre.
km if its speed is increased 
from its usual speed. Find the usual speed of the train.Solution: Let’s assume the usual speed of train as x km/hr Then, the increased speed of the train $= (x + 5) km/hr$ Using, speed = distance/ time Time taken by the train under usual speed to...
Answer- Because diamond has a refractive index of 2.42, the speed of light in it is reduced by a factor of 2.42 when compared to the speed of light in air. To put it another way, the speed of light...
Material mediumRefractive indexMaterial mediumRefractiveindexAir1.0003Canada Balsam1.53Ice1.31––Water1.33Rock salt1.54Alcohol1.36––Kerosene1.44Carbon disulphide1.63Fusedquartz1.46Denseflint...
. If the speed of the slow train is less than that of the fast train, find the speed of the two trainsSolution: Let’s consider the speed of the fast train as x km/hr Then, the speed of the slow train will be $= (x -10) km/hr$ Using, speed = distance/ time Time taken by the fast train to cover $200...
km, would have taken 
minutes less to travel the same distance if its speed were 
km/hr more. Find the original speed of the train.Solution: Let the original speed of train be x km/hr When increased by $5$, speed of the train $= (x + 5) km/hr$ Using, speed = distance/...
km/hr. It can go 
km upstream and 
km downstream in hours. Find the speed of the stream.Solution: Let the speed of stream be x km/hr Given, speed of boat in still water is $8km/hr$. So, speed of downstream $= (8 + x) km/hr$ And, speed of upstream $= (8 – x) km/hr$ Using, speed =...
MaterialmediumRefractive indexMaterial mediumRefractiveindexAir1.0003Canada Balsam1.53Ice1.31––Water1.33Rock salt1.54Alcohol1.36––Kerosene1.44Carbon disulphide1.63Fusedquartz1.46Denseflint...
Answer- According to the question, speed of light in vacuum (c) is 3 × 108 m/s And refractive index of glass (ng) is 1.50 Expression for the speed of light in the glass (v) is given by - v = Speed...
Answer- The light ray bends in the direction of normal. When a light ray passes through an optically rarer medium (low refractive index) and enters an optically denser medium (high refractive...
Answer- Expression for magnification produced by a spherical mirror is - m = (height of image)/(height of object) Also, m = -(Image distance)/(object distance) So, we can write - (height of...
Answer- According to the question, radius of curvature (R) is 32 cm Expression for radius of curvature of the spherical mirror = 2 × Focal length (f) R = 2f Or, f= R/2 f= 32 / 2 f = 16 Therefore, 16...
Answer- In cars and vehicles, a convex mirror is favoured as a rear-view mirror. This is because it provides a broader field of view, allowing the driver to see more of the traffic behind him....
Answer- Concave Mirror is a mirror that may magnify and construct an object's image.
Answer- According to the question, radius of curvature (R) is 20 cm Expression for radius of curvature of the spherical mirror is = 2 × Focal length (f) So, R = 2f Or, f= R/2 = 20 / 2 f = 10...
Solution: No. Justification: Consider 3q + 1as the positive integer, where q is a natural number. (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1, (where m being an integer equals to 3q2 +...
Solution: No, the square of any positive integer cannot be of the form 3m + 2 where m is a natural number Justification: Applying the Euclid’s division lemma, The positive integer ‘a' can be written...
Solution: Yes, the statement is true. Justification: Consider 2, 3, 4 as the three consecutive numbers \[\left( 2\text{ }\times \text{ }3\text{ }\times \text{ }4 \right)/6\text{ }=\text{ }24/6\text{...
Solution: Yes, the statement is true. Justification: Let ‘a’ and ‘a + 1’ be the two consecutive positive integers Applying the Euclid’s division lemma, We have, a = bq + r, where 0 ≤ r < b For...
Solution: No, every positive integer, where q is an integer cannot be of the form 4q + 2. Justification: All numbers in the form 4q + 2, where ‘q' is an integer, are even and not divisible by 4. For...
Answer- After reflecting from a concave mirror, light beams that are parallel to the principal axis converge at a certain location on the principal axis. The spot is the principal focus of the...
The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is (A) 13 (B) 65 (C) 875 (D) 1750 Solution: Option(A) 13 Explanation: According to the question, Find the...
If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is: (A) 4 (B) 2 (C) 1 (D) 3 Solution: Option(B) 2 Explanation: Let's calculate the HCF of 65 and 117, 117 = 1×65 +...
n2 – 1 is divisible by 8, if n is: (A) an integer (B) a natural number (C) an odd integer (D) an even integer Solution: Option(C) an odd integer Explanation: Let x = n2 – 1 In the equation...
For some integer q, every odd integer is of the form: (A) q (B) q +1 (C) 2g (D) 2q +1 Solution: Option(D) i.e. 2q+1 Explanation: Those integers which are not divisible by 2 are odd integers. As a...
For some integer m, every even integer is of the form: (A) m (B) m +1 (C) 2m (D) 2m+1 Solution: Option(C) 2m Explanation: Those integers which are divisible by 2 are even integers. As a result, it...
Think about the accompanying graph Volume of water that streams in t minutes from pipe \[=\text{ }t\times 0.5\pi \text{ }m^3\] Volume of water that streams in t minutes from pipe = \[t\times 0.5\pi...
a. Haploid dominantly b. With varying ploidy. c. A few haploid and a diploid d. Diploid Solution: Option (d) is the answer.
Solution: Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they...
No. of rooms12345678910No. of houses49222824128652 Solution: No. of roomsNo. of housesCumulative FrequencyLess than or equal to 144Less than or equal to 2913Less than or equal to 32235Less than or...
Height (in cm):160 – 162163 – 165166 – 168169 – 171172 – 174No of students:1511814212718 Find the average height of maximum number of students. Solution: Statistics is the discipline that...
![\[\mathbf{AB}\text{ }=\text{ }\mathbf{5}.\mathbf{6}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5bda87937897bc7402395a0e7497aa6b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{BC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cc05f2b9f3ba514ca9c9b35898d5010a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{BD}\text{ }=\text{ }\mathbf{3}.\mathbf{2}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cb291882d238e8010337f40ada99f74_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{AB}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-49cecb59e896652d21a82c8db7df1c60_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{AC}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8fcf95066978bb3a5c81e31da13743b6_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{BC}\text{ }=\text{ }\mathbf{12}\text{ }\mathbf{cm},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2442b612d2d4da073e2ebd36b6adccf8_l3.png "Rendered by QuickLaTeX.com")
Solution: Given: Δ ABC and AD bisects ∠A, meeting side BC at D. \[AB\text{ }=\text{ }5.6\text{ }cm,\] \[BC\text{ }=\text{ }6\text{ }cm,\] and \[BD\text{ }=\text{ }3.2\text{ }cm\]....
![\[\mathbf{24cm}\text{ }\mathbf{and}\text{ }\mathbf{10cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c85c0e710149fb9633b94c8c2abffd5a_l3.png "Rendered by QuickLaTeX.com")
Solution: Let ABCD be a rhombus and AC and BD be the diagonals of ABCD. So, AC = \[24cm\text{ }and\text{ }BD\text{ }=\text{ }10cm\] \[\] We know that diagonals of a rhombus bisect each other at...
![\[\mathbf{2a}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dc21b199c4adb0b51f7839a0d5967962_l3.png "Rendered by QuickLaTeX.com")
(i) AD = a\[\surd \mathbf{3}\] (ii) Area (∆ABC) = \[\surd \mathbf{3}\text{ }{{\mathbf{a}}^{\mathbf{2}}}\] Solution: (i) In ∆ABD and ∆ACD, we have∠ADB = ∠ADC = \[{{90}^{o}}\]AB = AC [Given]AD =...
![\[\mathbf{13cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0a3890f5045c3fc67ed8d0dc3fee80ac_l3.png "Rendered by QuickLaTeX.com")
\[\mathbf{5cm},\]find BC. Solution: Given, An isosceles triangle ABC, AB = AC = \[13cm,\text{ }AD\text{ }=\text{ }5cm\] Required to find: BC In ∆ ADB, by using Pythagoras theorem, we have...
![\[\mathbf{108c}{{\mathbf{m}}^{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aea3dbba23caac08c8a64b556bd738bb_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, ABCD is a square. And, F is the mid-point of AB. BE is one third of BC. Area of ∆ FBE = \[108c{{m}^{2}}\] Required to find: length of AC Let’s assume the sides of the square to be...
![\[\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{12}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{13}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-55595a46a936f27b67a56d78fd6d9f6b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{13}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd9682929f90290113c48dab79d05fcc_l3.png "Rendered by QuickLaTeX.com")
Solution: From the fig. \[AB\text{ }=\text{ }5cm,\text{ }BC\text{ }=\text{ }12\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }13\text{ }cm.\] Then, \[A{{C}^{2}}~=\text{ }A{{B}^{2}}~+\text{...
![\[.\text{ }\mathbf{4}.\mathbf{219}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e0e6c8ae5a8e273c9a6be45b04989009_l3.png "Rendered by QuickLaTeX.com")
Solution: We have, In ∆BAC, by Pythagoras theorem, we have \[\begin{array}{*{35}{l}} B{{C}^{2}}~=\text{ }A{{B}^{2}}~+\text{ }A{{C}^{2}} \\ \Rightarrow...
![\[\mathbf{9}\text{ }\mathbf{in}\text{ }\mathbf{and}\text{ }\mathbf{14}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5efcdbe8ed132a615044b2b9e1efbd4_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{12}\text{ }\mathbf{m},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-18666dfb8dc6e2f5280fce139ea621e6_l3.png "Rendered by QuickLaTeX.com")
Solution: Comparing with the figure, it’s given that AC = \[14\text{ }m,\text{ }DC\text{ }=\text{ }12m\text{ }and\text{ }ED\text{ }=\text{ }BC\text{ }=\text{ }9\text{ }m\] Construction: Draw...
![\[\mathbf{6}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0a07f6483f265e6e47c316121ac8cca7_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{8}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f5dd6c4c521acc9e7e7137b3036d18fb_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume the length of ladder to be, AD = BE = x m So, in ∆ACD, by Pythagoras theorem We have, \[\begin{array}{*{35}{l}} ...
![\[\mathbf{AC}\text{ }=\text{ }\mathbf{25}\text{ }\mathbf{cm},\text{ }\mathbf{BC}\text{ }=\text{ }\mathbf{14}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-49a79c647484ca1f07eb3412c6395529_l3.png "Rendered by QuickLaTeX.com")
Solution: Given, ∆ABC, AB = AC = \[25\text{ }cm\text{ }and\text{ }BC\text{ }=\text{ }14.\] \[\] In ∆ABD and ∆ACD, we see...
![\[\mathbf{6}\text{ }\mathbf{m}\text{ }\mathbf{and}\text{ }\mathbf{11}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ce51d1c0d04bc918a23933dbe2116e9a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{12}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5b1a778c62eb81bbb683b987a69b68d9_l3.png "Rendered by QuickLaTeX.com")
Solution: Let CD and AB be the poles of height \[11m\text{ }and\text{ }6m.\] Then, its seen that CP = \[11\text{ }\text{ }6\text{ }=\text{ }5m.\] From the figure, AP should be \[12m\] (given) In...
![\[\mathbf{17}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a8857d36c65c9db76f01d8b00e922493_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{15}\text{ }\mathbf{m}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f363f724150e591edf9182f4311ab3e8_l3.png "Rendered by QuickLaTeX.com")
Solution: In ∆ABC, by Pythagoras theorem \[\begin{array}{*{35}{l}} A{{B}^{2}}~+\text{ }B{{C}^{2}}~=\text{ }A{{C}^{2}} \\ \Rightarrow {{15}^{2}}~+\text{...
![\[\mathbf{15}\text{ }\mathbf{metres}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-712e26540491c89b51fc3bb96dfabd2f_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{8}\text{ }\mathbf{metres}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9d6db488112b88ddeb15d838748fedad_l3.png "Rendered by QuickLaTeX.com")
Solution: ...
\[\begin{array}{*{35}{l}} \left( \mathbf{i} \right)\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm},\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{24}\text{...
![\[\mathbf{3}\text{ }\mathbf{cm},\text{ }\mathbf{4}\text{ }\mathbf{cm},\text{ }\mathbf{and}\text{ }\mathbf{6}\text{ }\mathbf{cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4e7f08d95ac5afd7a7bf17c308701538_l3.png "Rendered by QuickLaTeX.com")
Solution: We have, Sides of triangle as \[\begin{array}{*{35}{l}} AB\text{ }=\text{ }3\text{ }cm \\ BC\text{ }=\text{ }4\text{ }cm \\ AC\text{...
$x:$$5$$6$$7$$8$$9$$f:$$4$$8$$14$$11$$3$ Solution: $x$$f$$fx$$5$$4$$20$$6$$8$$48$$7$$14$$98$$8$$11$$88$$9$$3$$27$$N=40$$\sum{fx=281}$ Mean $=\sum{fx/n=281/40}$ $\therefore $ Mean...
Solution: i. Structural: In deciduous trees, separation zone is formed at the base of the petiole. It is composed of a top layer(distal) and a bottom (proximal)layer. The cells in the top layer have...
![\[\mathbf{16}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ec8d9d2d731b4ca460e57ad46a988282_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{380}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cb340828e6c6101fbde249e54a9cd1cb_l3.png "Rendered by QuickLaTeX.com")
Solution: On dividing \[380\]by \[16\]we get,\[23\]as the quotient and\[12\]as the remainder. Now, since the LCM is not exactly divisible by the HCF its can be said that two numbers cannot have...
![\[\mathbf{84},\text{ }\mathbf{90}\text{ }\mathbf{and}\text{ }\mathbf{120}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ccfecca6ebaa7e44c50bda4ca8c75939_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{24},\text{ }\mathbf{15}\text{ }\mathbf{and}\text{ }\mathbf{36}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-94f859fd8a71046e251dab20e2d08dcb_l3.png "Rendered by QuickLaTeX.com")
Solution: First, Find the prime factors of the given integers: \[84,\text{ }90\text{ }and\text{ }120\] For, \[\begin{array}{*{35}{l}} ~84\text{...
![\[\mathbf{8},\text{ }\mathbf{9}\text{ }\mathbf{and}\text{ }\mathbf{25}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-78297b7f285a41777161d421fedfd300_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{40},\text{ }\mathbf{36}\text{ }\mathbf{and}\text{ }\mathbf{126}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17883cf51f0a9421203bd7e5576dc901_l3.png "Rendered by QuickLaTeX.com")
Solution: First, find the prime factors of the given integers: 8, 9 and 25 For, \[8\text{ }=\text{ }2\text{ }\times...
(i) 12, 15 and 21(ii) \[\mathbf{17},\text{ }\mathbf{23}\text{ }\mathbf{and}\text{ }\mathbf{29}\] Solution: First, find the prime factors of the given integers: \[12,\text{ }15\text{ }and\text{...
(i) \[\mathbf{26}\text{ and }\mathbf{91}\](ii) \[\mathbf{510}\text{ and }\mathbf{92}\] Solution: Given integers are\[\mathbf{26}\text{ and...
![\[\surd 5\text{ }+\text{ }\surd 3~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0efb8e4ff13d751cc9c194f8bc9b1b82_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[\surd 5\text{ }+\text{ }\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that\[\surd 5\text{ }+\text{ }\surd...
![\[\mathbf{2}\text{ }-\text{ }\mathbf{3}\surd \mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-492fc7220996be5495f8ab5970d52502_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[2\text{ }\text{ }3\surd 5\]is a rational number. Then, there exist co prime positive integers a and b such that \[2\text{ }\text{ }3\surd 5\text{...
![\[\mathbf{5}\text{ }-\text{ }\mathbf{2}\surd \mathbf{3}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4fe99f6574a18dae472d801d66331d39_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[5\text{ }\text{ }2\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that \[5\text{ }\text{ }2\surd 3\text{...
![\[\mathbf{4}\text{ }-\text{ }\mathbf{5}\surd \mathbf{2}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-728fc3aae21f5413024bcf441fffe566_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[4\text{ }\text{ }5\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[4\text{ }\text{ }5\surd 2\text{...
![\[\mathbf{3}\text{ }+\text{ }\surd \mathbf{2}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fc1c82b2f160b7a1dc310519dbde254a_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[3\text{ }+\text{ }\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[3\text{ }+\text{ }\surd 2=\text{...
![\[\mathbf{2}\text{ }-\text{ }\surd \mathbf{3}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d0a39d7c5e26f9ab889ba6084fa2705f_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[2\text{ }\text{ }\surd 3\] is a rational number. Then, there exist co prime positive integers a and b such that\[2\text{ }\text{ }\surd 3=\text{ }a/b\]...
![\[\mathbf{4}\text{ }+\text{ }\surd \mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e72e18bc9159e65a933a5182ce232b1e_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{5}\surd \mathbf{2}~~~~~~~~~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f66067211a2d0fa3d76358415a69dc68_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[4\text{ }+\text{ }\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[4\text{ }+\text{ }\surd 2\text{...
\[\mathbf{13}.\]\[~\mathbf{x}/\mathbf{a}\text{ }+\text{ }\mathbf{y}/\mathbf{b}\text{ }=\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{b}\] \[\mathbf{x}/{{\mathbf{a}}^{\mathbf{2}~}}+\text{...
![\[\mathbf{6}\text{ }+\text{ }\surd \mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3751ac6d34393366c041894ce911cbaa_l3.png "Rendered by QuickLaTeX.com")
Solution: Let’s assume on the contrary that \[6+\surd 2\] is a rational number. Then, there exist co prime positive integers a and b such that \[6\text{ }+\text{ }\surd 2\text{ }=\text{ }a/b\]...
Solution: The correct response is (b) lead acetate. Explanation: Because we need a compound that contains lead in order to obtain lead iodide, ammonium nitrate and potassium sulphate are ruled out....
Answer: option b Water contains two moles of hydrogen and one mole of water in a single mole. Because of this, the mole ratio of hydrogen to oxygen is 2:1.
Solution: Answer is option d) The elements ammonium and barium are being displaced from their respective salts, according to the data. As a result, we have a double displacement reaction. Because of...
Solution: Answer is option a) Silver halides, particularly silver chloride, decompose in the presence of sunlight, resulting in the formation of silver metal and a halogen gas (silver) (chlorine or...
Solution: The answer is (b) (ii) and (iii) In the presence of vigorous water reaction, solid calcium oxide transforms into calcium hydroxide, which is accompanied by the production of heat. This...
Solution: The answer is (b) (ii) only Sodium and barium are both displaced from their respective salts in this reaction, which is referred to as a double displacement reaction.
Solution: Answer is (a) KMnO4 is an oxidising agent, it oxidises FeSO4 Potassium permanganate (KMnO4) is used as an oxidising agent during this reaction. This is due to the presence of KMnO4, which...
Answer: Answer is (c) (i) and (iv) Exothermic processes result in the release of enormous amounts of heat – as a result, when water reacts with quick lime and when acid reacts with water, heat...
Answer: Option a) Exothermic processes result in the release of enormous amounts of heat – as a result when water reacts with quick lime and when an acid reacts with water, heat energy is...
Answer: Option c) In this reaction, oxygen reacts with water to form oxidised water. Because oxygen is being removed from water, the water's oxygen content is being reduced. Water is a source of...
Solution: Answer is (c) (i) and (iii) The reaction described here is a combination of displacement and redox reactions. The displacement reaction occurs when oxygen displaces hydrogen in ammonia,...
Answer: Answer is (d) Combustion of Liquefied Petroleum Gas (LPG) Because a new compound is formed after burning, combustion is always accompanied by a chemical change, which is irreversible in...
Answer: (c) $C_2 H_5 OH ( aq )+ CH_3 COOH (l) \longrightarrow CH_3 COO C_2 H_5( aq )+ H_2 O ( t )$It is a neutralisation reaction and double displacement reaction.(d) $C_{2} H_{4}( g )+3 O_{2}(g)...
Answer: (a) $N {2}( g )+3 H {2}( g )=\frac{\text { catalyst }}{773 K } 2 NH {2}( g )$ It is a combination reaction. (b) $NaOH ( aq )+ CH _{3} COOH ( aq ) \longrightarrow CH_3 COONa ( aq )+ H_2 O...
Answer: c) Aqueous potassium iodide solution is passed through a chlorine gas stream, resulting in the formation of potassium chloride solution and solid iodine.This is a reaction with a single...
Answer: a) Iron (III) oxide reacts with aluminium to form molten iron and aluminium oxide in the first step of the Thermit reaction. This is a reaction with a single displacement. Fe2O3 + 2Al →...
CaCO_3 (s) →xCaO(s) + CO_2 (g)$Solution: (c) Zn(s) + H2 SO4 (aq) → ZnSO4 (aq) + H2 (g) (d) CaCO3 (s) →heat→ CaO(s) + CO2 (g) x-(aq) y(g)x is heat
→
→
Answer: (a) Pb(NO3 ) 2 (aq) + 2KI(aq) → PbI2 (s) + 2KNO3 (aq) (b) Cu(s) + 2Ag NO3 (aq) → Cu(NO3 ) 2 (aq) + 2Ag(s) x(s), y (aq)x is 2Ag
Answer:An exothermic process is one that generates heat as a byproduct. A portion of this heat is transferred to the surrounding environment. Heat must be supplied to the system from the surrounding...
Answer: In a redox chemical reaction, a reducing agent is an element or compound that "donates" or "loses" an electron to an electron recipient as part of the reaction. NH3-AmmoniaH2O – Water
→
→
Answer: In a redox chemical reaction, a reducing agent is an element or compound that "donates" or "loses" an electron to an electron recipient as part of the reaction. CO-Carbon momnoxide$H_2$-...
→
→
Answer: According to chemistry, an oxidising agent, or oxidising agent, is a substance that has the ability to oxidise other substances, or to accept electrons from them, in other words, it is an...
→
→
Answer:According to chemistry, an oxidising agent, or oxidising agent, is a substance that has the ability to oxidise other substances, or to accept electrons from them, in other words, it is an...
→
→
Answer:According to chemistry, an oxidising agent, or oxidising agent, is a substance that has the ability to oxidise other substances, or to accept electrons from them, in other words, it is an...
Solution: (c) 2CuSO4+4Kl →$2K_2SO_4+Cu_2I_2+I_2$
Answer: (a) Na2CO3 + HCl → NaCl + NaHCO3 (b) NaHCO3 + HCl → NaCl + H2O + CO2
Solution: Double displacement reactions are chemical reactions in which one component of each of the reacting molecules is exchanged in order to form the products, as defined by the American...
Answer: When heated, ferrous sulfate decomposes into ferric oxide, sulphur dioxide, and sulphur trioxide, which are all toxic. Using this balanced equation, we can express the thermal decomposition...
Answer: During the course of the night, fireflies produce a chemical reaction within their bodies that allows them to glow. In the presence of an enzyme known as luciferase, oxygen reacts with...
Answer: The defense mechanism of plants prevents the fermentation of grapes on the plant. When grapes are plucked from the vine, they react with yeast to produce fermentation, which is the final...
Answer: a) Because silver is a member of the low reactive series of metals, there will be no reaction between it and dilute hydrogen chloride. b) Due to the fact that it is an exothermic reaction,...
Answer: c) Sodium is a highly reactive metal, and when it reacts with atmospheric oxygen, it produces an exothermic reaction, which raises the temperature of the surrounding environment. d) Hydrogen...
Answer: Calcium oxide is the chemical compound X. CaO is a compound that is widely used in the cement industry. Water treatment of CaO results in the formation of Ca(OH)2, which is alkaline in...
Answer: (c) Fe2O3 + 3CO + 2Fe + 3CO2 This is a redox reaction. (d) 2 H2S+O2 → 2 s + 2 H2O This is a replacement reaction.
Answer: (a) Pb(CH3COO)2 + 2HCI – PbCl2 + CH3COOH This is a Double Displacement reaction. (b) 2Na + 2C2H5OH + 2C2H5ONa+ H2 This is a Displacement reaction.
Answer: When exposed to sunlight, silver chloride decomposes into silver and chlorine gas, which are both toxic. As a result, silver chloride is stored in bottles that are dark in colour.
Answer: (c) 2Na(s) + S(s) — (Fuse) → Na2S(s) This is an example of a reaction known as a Combination reaction. (d) TiCI4 (1) + Mg(s) → Ti(s) + 2MgCl2 (s) There are several types of displacement...
Solution: (e) Cao(s) + SIO2(s) + CaSIO3(s) This reaction falls under the category of Displacement reactions (f) 2H2O2 (I) — UV → 2H2O(I) + O2 (g) This is a decomposition reaction.
(g) → 
(s) 
(g)Answer: (a) Mg(s) + Cl2(g) → MgCl2(s) A combination reaction, also known as a synthesis reaction, is a type of reaction in which two or more substances are combined. (b) 2HgO(s) — (Heat) → 2 Hg(I) +...
Solution: a) It is formed when magnesium ribbon is burned in oxygen with the emission of light and heat energy that magnesium oxide is formed.X is a compound with the chemical formula MgO. After...
Answer: Copper is more reactive than zinc because zinc is placed above hydrogen in the activity series of metals, whereas copper is placed below hydrogen in the activity series of metals. As a...
Solution: A reaction between silver and H2S in the atmosphere results in the formation of Silver Sulphide, which is a dark brown compound with a sulfuric acid smell. Corrosion is the term used to...
Answer: c) Thermal decomposition is the reaction that is taking place. d) Due to the fact that NO2 dissolves in water and forms an acidic solution, pH is less than 7. (pH range below 7).