Given bend conditions are: \[y2\text{ }=\text{ }4x\text{ }\ldots \text{ }.\text{ }\left( 1 \right)\text{ }and\text{ }x2\text{ }+\text{ }y2\text{ }\text{ }6x\text{ }+\text{ }1\text{ }=\text{ }0\text{...
Solution:
The given bends are \[y\text{ }=\text{ }4\text{ }\text{ }x2\text{ }\ldots \text{ }.\text{ }\left( I \right)\text{ }and\text{ }y\text{ }=\text{ }x2\text{ }\ldots \text{ }\left( ii \right)\] Also, we...
Solution:
Solution:
ACCORDING TO EQUATION OF CURVE, \[\surd x\text{ }+\text{ }\surd y\text{ }=\text{ }4\] Presently, let (x1, y1) be he required point on the bend SO , \[\surd x1\text{ }+\text{ }\surd y1\text{ }=\text{...
Solution:
Solution:
Solution:
Given bends are conditions of two circles, \[xy\text{ }=\text{ }4\text{ }\ldots \text{ }..\text{ }\left( I \right)\] and \[x2\text{ }+\text{ }y2\text{ }=\text{ }8\text{ }\ldots \text{ }.\text{...
It's seen that the given bends are condition of two circles. \[2x\text{ }=\text{ }y2\text{ }\ldots \text{ }..\text{ }\left( 1 \right)\] and \[2xy\text{ }=\text{ }k\text{ }\ldots \text{ }..\text{...
The surface of the adsorbent adsorbs the gases that are desirable in the process of adsorption. If an area already has other gases, it is saturated so it cannot adsorb the desired gases from the...
Chemisorption involves the formation of chemical bonds that require activating energy. That's why it's called activated adsorption.
How about we consider the space of the primary square \[A1\text{ }=\text{ }x2\] Furthermore, space of the subsequent square be \[A2\text{ }=\text{ }y2\] Presently, \[A1\text{ }=\text{ }x2\text{...
The soap solution at low concentration acts as a standard electrolytic solution. But in the higher concentrations it behaves like a colloid and forms micelles.
When the gelatin is mixed with gold sol, the gelatin forms a protective layer on the gold sol, thus preventing it from getting coagulated. So gold sol recovers by adding gelatin, which is a...
The Clouds hold charges as they are colloids. Silver iodide (AgI) electrolyte assists in the process of coagulation between colloidal water particles in a cloud. Thus, forming large raindrops that...
We should expect x to be the length of the block. In this way, the volume of the solid shape \[V\text{ }=\text{ }x3\text{ }\ldots \text{ }.\text{ }\left( 1 \right)\] Considering that, \[dV/dt\text{...
Given, \[L\text{ }=\text{ }200\left( 10\text{ }\text{ }t \right)2\] where L addresses the quantity of liters of water in the pool. On separating both the sides w.r.t, t, we get \[dL/dt\text{...
Let AB is the stature of streetlamp post and CD is the tallness of the man with the end goal that \[AB\text{ }=\text{ }5\left( 1/3 \right)\text{ }=\text{ }16/3\text{ }m\text{ }and\text{ }CD\text{...
(i) adsorption of coloured substance (ii) adsorption of solvent (iii) absorption and adsorption both of solvent (iv) absorption of solvent Correct Answers: (i) adsorption of coloured...
Given, The interior range $$ \[r\text{ }=\text{ }3\text{ }cm\] What's more, outside sweep \[R\text{ }=\text{ }r\text{ }+\text{ }r\text{ }=3.0005\text{ }cm\] \[r\text{ }=\text{ }3.0005\text{...
Solution: Anionic cleansers are utilized in toothpaste to clean teeth and structure reasonable froth. Eg: sodium or ammonium lauryl sulfate
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The correct option is (D) none of these Given R on the set \[\left\{ 1,\text{ }2,\text{ }3 \right\}\]be defined by \[R\text{ }=\text{ }\left\{ \left( 1,\text{ }2 \right) \right\}\] Therefore, its...
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The correct option is (D) \[5\] Given, set \[\mathbf{A}\text{ }=\text{ }\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3} \right\}\] Now, the number of equivalence relations as follows...
The correct option is (B) transitive but not symmetric Given aRb ⇒ a is brother of b. This does not mean b is also a brother of a as b can be a sister of a. Therefore, R is not symmetric. aRb ⇒ a is...
The correct option is (C) equivalence Given aRb, if a is congruent to b, ∀ a, b ∈ T. Then, we have aRa ⇒ a is congruent to a; which is always true. So, R is reflexive. Let aRb ⇒ a ~ b b ~ a bRa So,...
![\[\mathbf{a}\text{ }*\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{1}\text{ }+\text{ }\mathbf{ab},\forall \mathbf{a},\text{ }\mathbf{b}\in \mathbf{R}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2961f64055cadc8b7cae88fe60a1b709_l3.png "Rendered by QuickLaTeX.com")
(i) Given that * is a binary operation defined on R by \[\mathbf{a}\text{ }*\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{1}\text{ }+\text{ }\mathbf{ab},\forall \mathbf{a},\text{ }\mathbf{b}\in...
![\[\begin{array}{*{35}{l}} \left( \mathbf{i} \right)\text{ }\mathbf{a}\text{ }*\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{a}\text{ }\text{ }\mathbf{b}\forall \mathbf{a},\text{ }\mathbf{b}\in \mathbf{Q}\text{ }\left( \mathbf{ii} \right)\text{ }\mathbf{a}\text{ }*\text{ }\mathbf{b}\text{ }=\text{ }{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}~\forall \mathbf{a},\text{ }\mathbf{b}\in \mathbf{Q} \\ \left( \mathbf{iii} \right)\text{ }\mathbf{a}\text{ }*\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{ab}\forall \mathbf{a},\text{ }\mathbf{b}\in \mathbf{Q}\text{ }\left( \mathbf{iv} \right)\text{ }\mathbf{a}\text{ }*\text{ }\mathbf{b}\text{ }=\text{ }{{\left( \mathbf{a}\text{ }\text{ }\mathbf{b} \right)}^{\mathbf{2}}}~\forall \mathbf{a},\text{ }\mathbf{b}\in \mathbf{Q} \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0a1f3a790697fd211e1ec8a051a8dcce_l3.png "Rendered by QuickLaTeX.com")
Given that * is a binary operation defined on Q. (i) \[a\text{ }*\text{ }b\text{ }=\text{ }a\text{ }\text{ }b,\forall a,\text{ }b\in Q\] and \[b\text{ }*\text{ }a\text{ }=\text{ }b\text{ }\text{...
![\[\mathbf{f}~\left( \mathbf{x} \right)\text{ }=\text{ }{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{3x}\text{ }+\text{ }\mathbf{1},~\mathbf{g}~\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{2x}\text{ }\text{ }\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2dd0a70d016c1b6709d883a89deaa7c8_l3.png "Rendered by QuickLaTeX.com")
Given, \[\mathbf{f}~\left( \mathbf{x} \right)\text{ }=\text{ }{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{3x}\text{ }+\text{ }\mathbf{1},~\mathbf{g}~\left( \mathbf{x} \right)\text{ }=\text{...
![\[\mathbf{A}\text{ }=\text{ }\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots \text{ }\mathbf{9} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4bba703b875f44961cf66438db00c2aa_l3.png "Rendered by QuickLaTeX.com")
![\[A\text{ }\times A\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b2d867ff818b6c236e399d822f63f0a7_l3.png "Rendered by QuickLaTeX.com")
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Given, \[\mathbf{A}\text{ }=\text{ }\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots \text{ }\mathbf{9} \right\}\]and (a, b) R (c, d) if a + d = b + c for \[\left( a,\text{ }b...
Solution: (b) The compound that causes general stimulant activity on the focal sensory system has a place with the class of sedatives.
![\[\mathbf{x},\text{ }\mathbf{y}\in \mathbf{N}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-53bce6e1c6a20d27eb7981b3f2a0d9cc_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{10},\text{ }\mathbf{x},\text{ }\mathbf{y}\in \mathbf{N}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ac2a04bdc6b7e6a9e6cf59fc647f9a46_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{x}\text{ }+\text{ }\mathbf{4y}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{x},\text{ }\mathbf{y}\in \mathbf{N}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-066b5246906e9f67cf4ce6457424ca50_l3.png "Rendered by QuickLaTeX.com")
(i) Given, x is greater than y; \[\mathbf{x},\text{ }\mathbf{y}\in \mathbf{N}\] If \[\left( x,\text{ }x \right)\in R\], then \[x\text{ }>\text{ }x\], which is not true for any \[x\in N\]. Thus, R...
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![\[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{x}/\mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-60d315a329816d5d7d74e09a6424ff3b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{g}\left( \mathbf{x} \right)\text{ }=\text{ }\left| \mathbf{x} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7a2feecb05cc88418ae74c021378e4f5_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{h}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{x}\left| \mathbf{x} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e4e9b95c341051db94c9983ee99d99dc_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{k}\left( \mathbf{x} \right)\text{ }=\text{ }{{\mathbf{x}}^{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-531f812814bd08b1132646e938b5d74e_l3.png "Rendered by QuickLaTeX.com")
Given, \[A\text{ }=\text{ }\left[ 1,\text{ }1 \right]\] (i) \[f:\text{ }\left[ -1,\text{ }1 \right]\text{ }\to \text{ }\left[ -1,\text{ }1 \right],\text{ }f\text{ }\left( x \right)\text{ }=\text{...
![\[\mathbf{A}\text{ }=\text{ }\mathbf{R}\text{ }\text{ }\left\{ \mathbf{3} \right\},\text{ }\mathbf{B}\text{ }=\text{ }\mathbf{R}\text{ }\text{ }\left\{ \mathbf{1} \right\}.\text{ }\mathbf{Let}\text{ }\mathbf{f}\text{ }:\text{ }\mathbf{A}\text{ }\to \text{ }\mathbf{B}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4826e8120534bd03f6ed6670ce6482c3_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{f}\text{ }\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{x}\text{ }\text{ }\mathbf{2}/\text{ }\mathbf{x}\text{ }\text{ }\mathbf{3}\forall \mathbf{x}\in \mathbf{A}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3be59825d072da5199347321d3e97a56_l3.png "Rendered by QuickLaTeX.com")
According to the question, \[A\text{ }=\text{ }R\text{ }\text{ }\left\{ 3 \right\},\text{ }B\text{ }=\text{ }R\text{ }\text{ }\left\{ 1 \right\}\] And, f : A → B be defined by \[f\text{ }\left( x...
(i) Let f: N → N, be a mapping defined by \[f\text{ }\left( x \right)\text{ }=\text{ }{{x}^{2}}\] For \[f\text{ }({{x}_{1}})\text{ }=\text{ }f\text{ }({{x}_{2}})\] Then, \[{{x}_{1}}^{2}~=\text{...
Solution: (c) The most valuable grouping of medications for therapeutic physicists. It is based on sub-atomic targets. Target particles are typically biomolecules like carbs, lipids, proteins and...
![\[\mathbf{A}\text{ }=\text{ }\left\{ \mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1d0332e8f8738588de84e4d9d4c7882a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{B}\text{ }=\text{ }\left\{ \mathbf{2},\text{ }\mathbf{5},\text{ }\mathbf{6},\text{ }\mathbf{7} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5d67b0b282f80ff369f931d204458d4d_l3.png "Rendered by QuickLaTeX.com")
Given, \[\mathbf{A}\text{ }=\text{ }\left\{ \mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4} \right\}\],\[\mathbf{B}\text{ }=\text{ }\left\{ \mathbf{2},\text{ }\mathbf{5},\text{ }\mathbf{6},\text{...
![\[\mathbf{R}\text{ }=\text{ }\{\left( \mathbf{x},\text{ }\mathbf{y} \right):\text{ }\mathbf{x}\in \mathbf{N},\text{ }\mathbf{y}\in \mathbf{N},\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{41}\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-88df5ebd7ad6cdd01200d5d66cf2b418_l3.png "Rendered by QuickLaTeX.com")
Given function: \[\mathbf{R}\text{ }=\text{ }\{\left( \mathbf{x},\text{ }\mathbf{y} \right):\text{ }\mathbf{x}\in \mathbf{N},\text{ }\mathbf{y}\in \mathbf{N},\text{ }\mathbf{2x}\text{ }+\text{...
\[\left( 1.999 \right)5\text{ }=\text{ }\left( 2\text{ }\text{ }0.001 \right)5\] Let \[x\text{ }=\text{ }2\text{ }and\text{ }x\text{ }=\text{ }-\text{ }0.001\] Likewise, let \[y\text{ }=\text{ }x5\]...
question suggests, angle is \[\pi /3.\]
As per the inquiry, we have In this manner, the necessary point is \[\pi /3.\]
![\[\mathbf{sin}\text{ }(\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( \mathbf{0}.\mathbf{75} \right))\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-68c7d427817e899bb8b5a68d5aed6574_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{0}.\mathbf{75}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-446d3f09e70b6cad0293b7ed4c3da967_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{1}.\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0be31f8506cd6a9fb18570daae72de29_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{0}.\mathbf{96}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f0483be7dacb7fe556620b46cb0154b5_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{sin}\text{ }\mathbf{1}.\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0b77460703a89e6fbd25aab1d917b285_l3.png "Rendered by QuickLaTeX.com")
The correct option is (c) \[\mathbf{0}.\mathbf{96}\] We have, \[sin\text{ }(2\text{ }ta{{n}^{-1}}\left( 0.75 \right))\]
![\[\mathbf{cos}\text{ }(\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{2}/\mathbf{5}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{-\mathbf{1}}}~\mathbf{x})\text{ }=\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1cb547e2a81fe2ddca2b94ec20b7efdb_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{1}/\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-698ecc2591e4187b3b31451e4a8412ac_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{2}/\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5995ed954b2b025ca426324f5a89a8fe_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fd0385a423a02a2235879aff6bfece5b_l3.png "Rendered by QuickLaTeX.com")
The correct option is (b) \[\mathbf{2}/\mathbf{5}\] Given, \[cos\text{ }(si{{n}^{-1}}~2/5\text{ }+\text{ }co{{s}^{-1}}~x)\text{ }=\text{ }0\] So, this can be rewritten as \[\begin{array}{*{35}{l}}...
![\[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\sqrt{\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dfbf35cc7cb2d872feaf569485ef4904_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ \mathbf{1},\text{ }\mathbf{2} \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fcfcb28dd59583bc3e7a020a8cf5f451_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ -\mathbf{1},\text{ }\mathbf{1} \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b21f7411312e08e9e9c76b1a5369f211_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ \mathbf{0},\text{ }\mathbf{1} \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4951cedbbb831e919142a21d14151c6a_l3.png "Rendered by QuickLaTeX.com")
The correct option is (a) \[\left[ \mathbf{1},\text{ }\mathbf{2} \right]\] We know that, \[si{{n}^{-1}}~x\] is defined for \[x\in \left[ -1,\text{ }1 \right]\] So, f(x) =...
![\[\mathbf{cos}-\mathbf{1}\text{ }\left( \mathbf{2x}-\text{ }\text{ }\mathbf{1} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-99b8e6286b585837ce48c8b01132334e_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ \mathbf{0},\text{ }\mathbf{\pi } \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4f3b7c31075d2dae94fe9c4f60b1821a_l3.png "Rendered by QuickLaTeX.com")
The correct option is (a) \[\left[ \mathbf{0},\text{ }\mathbf{1} \right]\] Since, \[cos-1\text{ }x\] is defined for \[x\in \left[ -1,\text{ }1 \right]\] So, f(x) = \[\mathbf{cos}-\mathbf{1}\text{...
![\[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{cos}\text{ }\mathbf{33\pi }/\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aea705b6330621ecb7f543a1fea098c8_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{3\pi }/\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c2d1bda05bc0967141df29b8057cf78a_l3.png "Rendered by QuickLaTeX.com")
![\[-\mathbf{7\pi }/\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c3e656e247ec5c6b18a8ecf530d059fe_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{\pi }/\mathbf{10}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-47aaedd99c7dba30799394a11a516fb5_l3.png "Rendered by QuickLaTeX.com")
![\[-\mathbf{\pi }/\mathbf{10}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4b6678b39bfd2e446cdbc2e73b32692d_l3.png "Rendered by QuickLaTeX.com")
The correct option is (d) \[-\mathbf{\pi }/\mathbf{10}\]
How about we believe P to be any point where the two streets are leaned at a point of \[45o.\] Presently, two men An and B are moving along the streets PA and PB separately with same speed 'V'....
Speed of the kite(V) \[=\text{ }10\text{ }m/s\] Leave FD alone the tallness of the kite and AB be the stature of the kite and AB be the tallness of the kid. Presently, let AF \[=\text{ }x\text{ }m\]...
![\[\mathbf{3}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}}}~\mathbf{x}\text{ }=\text{ }\mathbf{\pi }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f24489a3235ab60a7d1ead918fee879b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0313c9365144662794ca80515dcd72ce_l3.png "Rendered by QuickLaTeX.com")
![\[-\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-96158281230e100426cf2f540c6264ad_l3.png "Rendered by QuickLaTeX.com")
The correct option is (b) \[\mathbf{1}\] Given, \[\mathbf{3}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}}}~\mathbf{x}\text{...
![\[\mathbf{cose}{{\mathbf{c}}^{-\mathbf{1}}}~\mathbf{x}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6d1eb220a0e4116384d174374ab9174c_l3.png "Rendered by QuickLaTeX.com")
![\[\left( -\mathbf{\pi }/\mathbf{2},\text{ }\mathbf{\pi }/\mathbf{2} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-04e628daf78abc2132895875d18d553a_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ \mathbf{0},\text{ }\mathbf{\pi } \right]\text{ }\text{ }\left\{ \mathbf{\pi }/\mathbf{2} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1095ec138b1c83380bf46ca1d2d9c985_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ -\mathbf{\pi }/\mathbf{2},\text{ }\mathbf{\pi }/\mathbf{2} \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5354ebba5b1885dfedd7d083862690a1_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ -\mathbf{\pi }/\mathbf{2},\text{ }\mathbf{\pi }/\mathbf{2} \right]\text{ }\text{ }\left\{ \mathbf{0} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6730dd1a41f4218ce47d3ee43e4842e8_l3.png "Rendered by QuickLaTeX.com")
The correct option is (d) \[\left[ -\mathbf{\pi }/\mathbf{2},\text{ }\mathbf{\pi }/\mathbf{2} \right]\text{ }\text{ }\left\{ \mathbf{0} \right\}\] According to the principal branch of...
We realize that the space of circle, \[A\text{ }=\text{ }\pi r2\] , where r = span of the circle Furthermore, \[edge\text{ }=\text{ }2\pi r\] As indicated by the inquiry, we have Subsequently, it's...
Given, a round ball of salt Then, at that point, the volume of ball \[V\text{ }=\text{ }4/3\text{ }\pi r3\] where r = sweep of the ball Presently, as per the inquiry we have \[dV/dt\propto S\] ,...
Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in the figure (b). Given that the input...
Answer: When signal 20sinωt provides an input voltage less than 5 volt (because after 5V, the diode will receive a positive voltage at its P-junction), the diode will be in reverse bias, and the...
Answer: As $ {{V}_{BE}}=0 $, the potential drop across $ {{R}_{B}} $ is 10 Volts. $ {{I}_{B}}=\frac{10}{400\times {{10}^{3}}}=25\mu A $ Now, as $ {{V}_{CE}}=0 $, The potenial drop across $...
Answer: Because the diode in that branch is reverse biased, I3 is zero. I3 = 0 We know that the resistance in AB = 150 ohms and the resistance in EF = 150 ohms and AB is parallel to EF Therefore,...
Answer: According to the question, Power = 1W and Zener breakdown voltage = 5V We are given that the minimum voltage = 3V and the maximum voltage = 7V $ {{I}_{\max }}=\frac{P}{{{V}_{z}}}=\frac{1}{5}...
Answer: The given circuit is AND and the truth table is: A B X = A.B 0 0 0 0 1 0 1 0 0 1 1 1
Answer: The bandgap in an elemental conductor is such that the emissions are infrared rather than visible. Elemental semiconductors cannot be used to make visible LEDs. This is because the energy...
Answer: a) There is only one input and only one output b) By making use of the Boolean expression c) Realization of NOT gate d) Making use of the truth table The NOT gate is a single-input,...
Answer: When an automobile approaches the gate, one or both gates are open. As a result, the OR gate produces the required result. The following is a truth table for the same: A B Y = A + B 0 0 0 0...
Answer: From the circuit diagram, we have IbR1 + Vbe = Vbb Expression for the base current gives us Ib = Vbb – Vbe/R1 It can be seen that Ib is inversely proportional to R1 The reading of the...
Answer: According to the question, the wavelength is: λ = 6000 Ao = 6000 × 10-10m Expression for the energy of the light photon is as follows: E = hc/ λ Upon substituting the values of h, c and...
(ii) What does the point P in the figure (a) represent? (iii) What does the points P and Q in the figure (b) represent? Answer: (i) Figure a) depicts the Zener diode's characteristics, whereas...
Answer: The DC supply is connected to the CE transistor amplifier to provide energy to the signal. As a result, the CE configuration amplifier has a significant power gain. The extra power necessary...
(i) if dc supply voltage is 10V? (ii) if dc supply voltage is 5V? Answer: According to the question, we can write Voltage gain in X = vx = 10 Voltage gain in Y = vy = 20 Voltage gain in Z = vz = 30...
The waveform formed across the resistance diode conducts when the diode is forward biassed in the given circuit, hence the output will be only when the input +1V is between t1 and t2. As a result,...
Answer: We can't use a voltmeter to detect the potential barrier across a p-n junction because the voltmeter's resistance must be very high relative to the junction resistance, which is almost...
Answer: If there is no energy gap between the conduction and valence bands in an atom's energy band diagram, the material will conduct current. This energy gap narrows when the material transitions...
Answer: Because the size of the dopant must be compatible with the semiconductor and they must form covalent bonds. The elemental dopants for silicon or germanium are usually chosen from groups XIII...
![\[\mathbf{cos}-\mathbf{1}\text{ }\mathbf{x}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d8e6ba03fbd43d222c09e6d7ff5e6405_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{0},\text{ }\mathbf{\pi } \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-304afae725f16f5826abffbd7d0e3505_l3.png "Rendered by QuickLaTeX.com")
![\[\left[ \mathbf{0}.\text{ }\mathbf{\pi } \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-96074c13411cdba9487f8ef3abfc6ff9_l3.png "Rendered by QuickLaTeX.com")
The correct answer is (c) \[\left[ \mathbf{0}.\text{ }\mathbf{\pi } \right]\] According to the principal value branch \[co{{s}^{-1}}~x\] is \[\left[ 0,\text{ }\pi \right]\].
![\[{{\mathbf{a}}_{\mathbf{1}}},\text{ }{{\mathbf{a}}_{\mathbf{2}}},\text{ }{{\mathbf{a}}_{\mathbf{3}}},\text{ }\ldots .,\text{ }{{\mathbf{a}}_{\mathbf{n}}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f0761db9221d8c4a34fb1a319790ab96_l3.png "Rendered by QuickLaTeX.com")
Given \[{{\mathbf{a}}_{\mathbf{1}}},\text{ }{{\mathbf{a}}_{\mathbf{2}}},\text{ }{{\mathbf{a}}_{\mathbf{3}}},\text{ }\ldots .,\text{ }{{\mathbf{a}}_{\mathbf{n}}}~\] is an arithmetic progression with...
(a) large velocity of the minority charge carriers if the doping concentration is small (b) large velocity of the minority charge carriers if the doping concentration is large (c) strong electric...
(a) RL should be increased (b) input frequency should be decreased (c) input frequency should be increased (d) capacitors with high capacitance should be used Answer: The correct options are: (a) RL...
![\[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~(\mathbf{1}/\mathbf{2}\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{4})\text{ }=\text{ }\left( \mathbf{4}\text{ }\text{ }\surd \mathbf{7} \right)/\text{ }\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dccff6f472609b4472e4046132f40b0b_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{4}\text{ }+\text{ }\surd \mathbf{7} \right)/\text{ }\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d3c7d7e4f6ad4b104e1c211168af33a9_l3.png "Rendered by QuickLaTeX.com")
We have, \[ta{{n}^{-1}}\left( 1/2\text{ }si{{n}^{-1}}~3/4 \right)\] Let, \[~{\scriptscriptstyle 1\!/\!{ }_2}\text{ }si{{n}^{-1}}~{\scriptscriptstyle 3\!/\!{ }_4}\text{ }=\text{ }\theta \] or,...
![\[\mathbf{4}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{5}\text{ }\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{239}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1c868672679244103bf531dc883f771b_l3.png "Rendered by QuickLaTeX.com")
\[\begin{array}{*{35}{l}} 4\text{ }ta{{n}^{-1}}~1/5\text{ }\text{ }ta{{n}^{-1}}~1/239 \\ =\text{ }2\text{ }(ta{{n}^{-1}}~1/5)\text{ }\text{ }ta{{n}^{-1}}~1/239 \\ \end{array}\] \[=\text{ }2\text{...
(a) The current in and voltage across the Zenor remains fixed (b) The current through the series Resistance (Rs ) changes (c) The Zener resistance is constant (d) The resistance offered by the Zener...
![\[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{4}\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{2}/\mathbf{9}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\surd \mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17821387054a6129ddeeec9c8d470549_l3.png "Rendered by QuickLaTeX.com")
Taking the LHS, \[ta{{n}^{-1}}~1/4\text{ }+\text{ }ta{{n}^{-1}}~2/9\] = RHS – Hence Proved
(a) there are no mobile charges (b) equal number of holes and electrons exist, making the region neutral (c) recombination of holes and electrons has taken place (d) immobile charged ions exist...
(a) The emitter current will be 8 mA (b) The emitter current will be 10.53 mA (c) The base current will be 0.53 mA (d) The base current will be 2 mA Answer: The correct options are: (b) The emitter...
![\[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{5}/\mathbf{13}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\text{ }=\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{63}/\mathbf{16}\text{ }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9f862e2ddc16f49b0d8cb191b1265cde_l3.png "Rendered by QuickLaTeX.com")
Solution: Here, \[si{{n}^{-1}}~5/13\text{ }=\text{ }ta{{n}^{-1}}~5/12\] And, \[co{{s}^{-1}}~3/5\text{ }=\text{ }ta{{n}^{-1}}~4/3\] Taking the L.H.S, we have Thus, L.H.S = R.H.S – Hence Proved...
(a) At Vi = 0.4V, transistor is in active state (b) At Vi = 1V, it can be used as an amplifier (c) At Vi = 0. .5V, it can be used as a switch turned off (d) At Vi = 2.5V, it can be used as a switch...
![\[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{8}/\mathbf{17}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\text{ }=\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{77}/\mathbf{85}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4a29de9d312057185353faf3423c5473_l3.png "Rendered by QuickLaTeX.com")
Taking the L.H.S, = \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{8}/\mathbf{17}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\] = tan-1 8/15 + tan-1 3/4 – Hence...
![\[\mathbf{co}{{\mathbf{s}}^{-\mathbf{1}}}~\left[ \mathbf{3}/\mathbf{5}\text{ }\mathbf{cos}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{4}/\mathbf{5}\text{ }\mathbf{sin}\text{ }\mathbf{x} \right],\text{ }\mathbf{x}\in \left[ -\mathbf{3\pi }/\mathbf{4},\text{ }\mathbf{\pi }/\mathbf{4} \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d850eeacf1c4cdcc74e478df322d0565_l3.png "Rendered by QuickLaTeX.com")
We have, \[\mathbf{co}{{\mathbf{s}}^{-\mathbf{1}}}~\left[ \mathbf{3}/\mathbf{5}\text{ }\mathbf{cos}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{4}/\mathbf{5}\text{ }\mathbf{sin}\text{ }\mathbf{x}...
Taking L.H.S, = R.H.S – Hence Proved
![\[\mathbf{cos}\text{ }(\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}~}}\mathbf{x})\text{ }=\text{ }\mathbf{sin}\text{ }(\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}~}}\mathbf{3}/\mathbf{4})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6cd380c08ad8ec5642d29a05211ae34a_l3.png "Rendered by QuickLaTeX.com")
Given equation, \[\mathbf{cos}\text{ }(\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}~}}\mathbf{x})\text{ }=\text{ }\mathbf{sin}\text{ }(\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}~}}\mathbf{3}/\mathbf{4})\]...
![\[\mathbf{cos}\text{ }\left( \mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{7} \right)\text{ }=\text{ }\mathbf{sin}\text{ }(\mathbf{4}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{3}).\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c138d201ed7e18941451df2a61b7d942_l3.png "Rendered by QuickLaTeX.com")
Taking L.H.S, we have Thus, L.H.S = R.H.S – Hence proved
![\[\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( \mathbf{cos}\text{ }\mathbf{\theta } \right)\text{ }=\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\left( \mathbf{2}\text{ }\mathbf{cosec}\text{ }\mathbf{\theta } \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d2c7880428f4dda9d218ce3ce1285562_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{\theta }\text{ }=\text{ }\mathbf{\pi }/\mathbf{4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8f04b641eff03fb75d5c846a8eb665ad_l3.png "Rendered by QuickLaTeX.com")
Given, \[\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( \mathbf{cos}\text{ }\mathbf{\theta } \right)\text{ }=\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\left( \mathbf{2}\text{...
![\[\mathbf{sin}\text{ }(\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{3})\text{ }+\text{ }\mathbf{cos}\text{ }(\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{2}\surd \mathbf{2})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5540e2f416552b974588e8588acb6daf_l3.png "Rendered by QuickLaTeX.com")
Given expression, \[\mathbf{sin}\text{ }(\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{3})\text{ }+\text{ }\mathbf{cos}\text{...
Given equation, Hence, the real solutions of the given trigonometric equation are \[0\] and \[-1\].
![\[\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( -\mathbf{3} \right)\text{ }=\text{ }\text{ }\mathbf{\pi }/\mathbf{2}\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( -\mathbf{4}/\mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1bf1fa38e67e8dd0a1d86ec247fb01c5_l3.png "Rendered by QuickLaTeX.com")
Taking L.H.S = \[2\text{ }ta{{n}^{-1}}\left( -3 \right)\text{ }=\text{ }-2\text{ }ta{{n}^{-1}}~3\text{ }(\because ta{{n}^{-1}}~\left( -x \right)\text{ }=\text{ }\text{ }ta{{n}^{-1}}~x)\text{ }\in...
![\[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}~}}\left( \mathbf{tan}\text{ }\mathbf{2\pi }/\mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d9347e59b88d99bd217f83bd865f0408_l3.png "Rendered by QuickLaTeX.com")
We know that, \[ta{{n}^{-1}}~tan\text{ }x\text{ }=\text{ }x,\text{ }x\in \left( -\pi /2,\text{ }\pi /2 \right)\]
![\[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\left( -\mathbf{1}/\surd \mathbf{3} \right)\text{ }+\text{ }\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}}}\left( \mathbf{1}/\surd \mathbf{3} \right)\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( \mathbf{sin}\text{ }\left( -\mathbf{\pi }/\mathbf{2} \right) \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-648eb67f374bfb1b0ea85a535aa93285_l3.png "Rendered by QuickLaTeX.com")
According to the question, \[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\left( -\mathbf{1}/\surd \mathbf{3} \right)\text{ }+\text{ }\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}}}\left( \mathbf{1}/\surd...
![\[\mathbf{cot}\text{ }(\mathbf{\pi }/\mathbf{4}\text{ }\text{ }\mathbf{-2}\text{ }\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}}}~\mathbf{3})\text{ }=\text{ }\mathbf{7}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-158939f2ec8b756711a45b4e1f072354_l3.png "Rendered by QuickLaTeX.com")
According to the question, L.H.S = R.H.S – Hence Proved
![\[\mathbf{cos}[\mathbf{co}{{\mathbf{s}}^{-\mathbf{1}}}\left( -\surd \mathbf{3}/\mathbf{2} \right)\text{ }+\text{ }\mathbf{\pi }/\mathbf{6}]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-129f4b73ad8a452af907b31ca9d311f8_l3.png "Rendered by QuickLaTeX.com")
(a) Electrons crossover from emitter to collector (b) Holes move from base to collector (c) Electrons move from emitter to base (d) Electrons from emitter move out of base without going to the...
(a) electrons move from lower energy level to higher energy level in the conduction band (b) electrons move from higher energy level to lower energy level in the conduction band (c) holes in the...
(a) A B E 0 0 1 0 1 0 1 0 1 1 1 0 (b) A B E 0 0 1 0 1 0 1 0 0 1 1 1 (c) A B E 0 0 0 0 1 1 1 0 0 1 1 1 (d) A B E 0 0 0 0 1 1 1 0 1 1 1 0 Answer: The correct option is (c) A B E 0 0 0 0 1 1 1 0 0 1 1...
(a) 1.3 V (b) 2.3 V (c) 0 (d) 0.5 V Answer: The correct option is (b) 2.3 V Explanation: Let V represent the potential difference between A and B, then we can write: V - 0.3 = (5 +5) x 103 x (0.2 x...
(a) would be zero at all times (b) would be like a half-wave rectifier with positive cycles in output (c) would be like a half-wave rectifier with negative cycles in output (d) would be like that of...
![\[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\left[ \mathbf{tan}\text{ }\left( \mathbf{5\pi }/\mathbf{6} \right) \right]\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{-\mathbf{1}}}~\left[ \mathbf{cos}\text{ }\left( \mathbf{13\pi }/\mathbf{6} \right) \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5cebef4c3e42c84da0e476a637feb0b1_l3.png "Rendered by QuickLaTeX.com")
We know that, \[ta{{n}^{-1}}~tan\text{ }x\text{ }=\text{ }x,\text{ }x\in \left( -\pi /2,\text{ }\pi /2 \right)\] And, here \[ta{{n}^{-1}}~tan\text{ }\left( 5\pi /6 \right)\text{ }=\text{ }5\pi...
(a) an anti-particle of the electron (b) a vacancy created when an electron leaves a covalent bond (c) absence of free electrons (d) an artificially created particle Answer: The correct option is...
(a) 220V (b) 110V (c) 0V (d) 220 √2 V Answer: The correct option is (d) 220 √2 V Explanation: V(max) will be the potential difference across the capacitor. It is given by: ${{V}_{\max...
(a) D1 is forward biased and D2 is reverse biased and hence current flows from A to B (b) D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice versa (c) D1...
(a) 1 and 3 both correspond to forward bias of junction (b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction (c) 1 corresponds to forward bias and 3...
(a) number density of free current carriers increases (b) relaxation time increases (c) both number density of carriers and relaxation time increase (d) number density of current carriers...
![\[\forall \mathbf{a},\text{ }\mathbf{b}\in \mathbf{Z}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ad8e1c13fb20f78579c0c903ffbb8c13_l3.png "Rendered by QuickLaTeX.com")
Given \[\forall \mathbf{a},\text{ }\mathbf{b}\in \mathbf{Z}\], aRb if and only if a – b is divisible by n. Now, for \[aRa\Rightarrow \left( a\text{ }\text{ }a \right)\]is divisible by n, which is...
![\[\mathbf{f}:\text{ }\mathbf{R}\text{ }\to \text{ }\mathbf{R}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd6ed4993146108e6223c8db2541e3f7_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{1}/\left( \mathbf{2}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{x} \right)\forall \mathbf{x}\in \mathbf{R}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-84cddfb14d8332eb9da79af0df936940_l3.png "Rendered by QuickLaTeX.com")
According to the question, \[\begin{array}{*{35}{l}} f\left( x \right)\text{ }=\text{ }1/\left( 2\text{ }\text{ }cos\text{ }x \right)\forall x\in R \\ Let\text{ }y\text{ }=\text{ }1/\left( 2\text{...
![\[\mathbf{f}:\text{ }\mathbf{A}\text{ }\to \text{ }\mathbf{B}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a65b4f87417f32d09792148a75d456c6_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{g}:\text{ }\mathbf{B}\text{ }\to \text{ }\mathbf{A}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1fe8f0a4ce8e1b942287832dc9c329ac_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{g}\text{ }\mathbf{o}\text{ }\mathbf{f}\text{ }=\text{ }{{\mathbf{I}}_{\mathbf{A}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2822f57295d364ff6b54fcd63ddbadb5_l3.png "Rendered by QuickLaTeX.com")
Given, \[\mathbf{f}:\text{ }\mathbf{A}\text{ }\to \text{ }\mathbf{B}\]and \[\mathbf{g}:\text{ }\mathbf{B}\text{ }\to \text{ }\mathbf{A}\] satisfy \[\mathbf{g}\text{ }\mathbf{o}\text{...
![\[\mathbf{X}\text{ }=\text{ }\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-00a236a7e355a7d670dcb8476df3ab87_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{Y}\text{ }=\text{ }\left\{ \mathbf{4},\text{ }\mathbf{5} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-213326489477be96ad3cd80e2d540705_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{X}\text{ }\times \text{ }\mathbf{Y}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f014b536a97f4ce864b7dfedf28891d9_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{f}\text{ }=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{4} \right),\text{ }\left( \mathbf{1},\text{ }\mathbf{5} \right),\text{ }\left( \mathbf{2},\text{ }\mathbf{4} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{5} \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6691f233314673e77e22007af106b401_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{g}\text{ }=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{4} \right),\text{ }\left( \mathbf{2},\text{ }\mathbf{4} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{4} \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5042e575b4ccda6cf2e48eaa227a27b8_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{h}\text{ }=\text{ }\left\{ \left( \mathbf{1},\mathbf{4} \right),\text{ }\left( \mathbf{2},\text{ }\mathbf{5} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{5} \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8b578f4fddc68e8e0e586ca65111a21a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{k}\text{ }=\text{ }\left\{ \left( \mathbf{1},\mathbf{4} \right),\text{ }\left( \mathbf{2},\text{ }\mathbf{5} \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-345e65558e5e27c048ec049113afadcd_l3.png "Rendered by QuickLaTeX.com")
Solution: From the question, \[\mathbf{X}\text{ }=\text{ }\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3} \right\}\]and \[\mathbf{Y}\text{ }=\text{ }\left\{ \mathbf{4},\text{ }\mathbf{5}...
![\[\forall \mathbf{x}\in \mathbf{R}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-93681e8b3dafd096e20967b3836271fa_l3.png "Rendered by QuickLaTeX.com")
We have, \[f:\text{ }R\text{ }\to \text{ }R\], \[f\left( x \right)\text{ }=\text{ }cos\text{ }x\] Now, \[\begin{array}{*{35}{l}} f\text{ }({{x}_{1}})\text{ }=\text{ }f\text{ }({{x}_{2}}) \\...
![\[\mathbf{f}:\text{ }\mathbf{C}\text{ }\to \text{ }\mathbf{R}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b91d0e5e712c414bc7e31733b22d22a9_l3.png "Rendered by QuickLaTeX.com")
![\[f\left( \mathbf{z} \right)\text{ }=\text{ }\left| \mathbf{z} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f825948db38f9f56374d6b1effa2f56c_l3.png "Rendered by QuickLaTeX.com")
![\[\forall \mathbf{z}\in \mathbf{C}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a74567bb0e077aca3e2b1f592c075bd3_l3.png "Rendered by QuickLaTeX.com")
According to the question, \[\mathbf{f}:\text{ }\mathbf{C}\text{ }\to \text{ }\mathbf{R}\] such that \[f\left( \mathbf{z} \right)\text{ }=\text{ }\left| \mathbf{z} \right|\], \[\forall \mathbf{z}\in...
![\[\mathbf{f}\text{ }=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{2} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{5} \right),\text{ }\left( \mathbf{4},\text{ }\mathbf{1} \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-58479171ac85089e2b7c869f6a70f385_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{g}\text{ }=\text{ }\left\{ \left( \mathbf{2},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{5},\text{ }\mathbf{1} \right),\text{ }\left( \mathbf{1},\text{ }\mathbf{3} \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6bae4d20c2ddb3f71bbf7e1016b46c25_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{f}\text{ }\mathbf{o}\text{ }\mathbf{g}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-079b2a9780de7bb51c6a828ea8162b7e_l3.png "Rendered by QuickLaTeX.com")
From the question, \[\mathbf{f}\text{ }=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{2} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{5} \right),\text{ }\left( \mathbf{4},\text{ }\mathbf{1}...
(i) from the question, {(x, y): x is a person, y is the mother of x} So we can say that each person ‘x’ has only one biological mother. Therefore, the above set of ordered pairs make a function. If...
![\[\mathbf{g}\text{ }=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{1} \right),\text{ }\left( \mathbf{2},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{5} \right),\text{ }\left( \mathbf{4},\text{ }\mathbf{7} \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d072d1f94a4403000866fa1969963f5e_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{g}\text{ }\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{\alpha x}\text{ }+\text{ }\mathbf{\beta }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e132af380935f39045049cfed6bc479e_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{\alpha }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2290b3b6ebb8da39f941931b782a55ee_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{\beta }\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-24941173bdd7b0c67a5c814c4779615f_l3.png "Rendered by QuickLaTeX.com")
From the question, \[\mathbf{g}\text{ }=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{1} \right),\text{ }\left( \mathbf{2},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{5}...
![\[\mathbf{f}\text{ }\left( \mathbf{x} \right)\text{ }=\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3x}\text{ }+\text{ }\mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-995851d8d7836e7b40ef98cf0bd42f1a_l3.png "Rendered by QuickLaTeX.com")
Given, \[\mathbf{f}\text{ }\left( \mathbf{x} \right)\text{ }=\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3x}\text{ }+\text{ }\mathbf{2}\] Then, \[\begin{array}{*{35}{l}} f\text{ }\left(...
![\[\mathbf{f}{{~}^{\mathbf{1}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-767f1094114853260386f6d809284ad0_l3.png "Rendered by QuickLaTeX.com")
According to the question, A = {a, b, c, d} and f = {(a, b), (b, d), (c, a), (d, c)} So, \[\mathbf{f}{{~}^{\mathbf{1}}}\] = {(b, a), (d, b), (a, c), (c, d)}
![\[\mathbf{2x}\text{ }\text{ }\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-022296614ad7968c0c3b37dedea5273e_l3.png "Rendered by QuickLaTeX.com")
According to the question the function, f (x) = \[\mathbf{2x}\text{ }\text{ }\mathbf{3}\], \[\forall \mathbf{x}\in \mathbf{R}\] Let us consider \[y\text{ }=\text{ }2x\text{ }\text{ }3\] \[x\text{...
![\[\mathbf{g}:\text{ }\mathbf{R}\text{ }\to \text{ }\mathbf{R}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2a6e4f932765ad5e387a51e3c62eeb38_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{2x}\text{ }+\text{ }\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-16fd72fb9940c8b247bd1589bb6b4d2a_l3.png "Rendered by QuickLaTeX.com")
![\[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e52890f89dba1d3f978c6f41597f9bd7_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{g}\text{ }\mathbf{o}\text{ }\mathbf{f}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-446f4667cf4569e1d8763649734e7995_l3.png "Rendered by QuickLaTeX.com")
According to the question, f(x) = \[\mathbf{2x}\text{ }+\text{ }\mathbf{1}\]and g (x) = \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{2}\], \[\forall \mathbf{x}\in \mathbf{R}\] Thus, \[g\text{...
![\[\sqrt{(\mathbf{25}\text{ }\text{ }{{\mathbf{x}}^{\mathbf{2}}})}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-653a123eb2d1b6788fe0e3cb5438b845_l3.png "Rendered by QuickLaTeX.com")
According to the question, f(x) = \[\sqrt{(\mathbf{25}\text{ }\text{ }{{\mathbf{x}}^{\mathbf{2}}})}\] The function is defined if \[25\text{ }\text{ }{{x}^{2}}~\ge \text{ }0\] So, \[{{x}^{2}}~\le...
According to the given question the relation, R = {(a, a), (b, c), (a, b)} Therefore, to make R as reflexive we should add (b, b) and (c, c) to R. Also, to make R as transitive we should add (a, c)...
In ortho-nitrophenol, there is an electron-withdrawing group NO2 in the ortho position, which increases the acidic strength and makes it more acidic. There is an electron releasing group in o-cresol...
Because Butan-2-ol possesses a chiral carbon atom, it is the only one that displays optical activity.
Ethanol water phenol ethanol ethanol ethanol ethanol ethanol ethanol ethanol ethanol Because it forms the phenoxide ion after deprotonation and is stabilized by resonance, phenol is more acidic....
is modified to 
Calculate in such a case, the ground state energy of an 
atom, if 
Case 1: When $\varepsilon=0.1, \mathrm{R_O}=1 \mathrm{~A}$ $\mathrm R_1=8 \times 10^{-11}$ $\mathrm{M}=0.08 \mathrm{~nm}$ Velocity at ground level is given as $\mathrm v_1=1.44 \times 10^{6}...
Auger electron emitted by Chromium by absorbing the energy from an 
to 
transition.The energy in the nth state can be calculated as, $-\operatorname{Rch} Z^{2} / n^{2}=-13.6 z^{2} / n^{2} e V$ Where $R=$ Rydberg constant and $\mathrm{Z}=24$ Energy released is given as $\left(13.6...
(i) Consider the nucleus of a H atom as a point charge electron circling about it at a speed of $v$ and a radius of $r_A$ The Coulombian force acts as a centrifugal force, causing the nucleus to...
and 
. This is because the wavelength of transition depends to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass 
, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here 
where 
is the nuclear mass and 
is the electronic mass. Estimate the percentage difference in wavelength for the 1 st line of the Lyman series in and . (Mass of nucleus is 
, Mass of nucleus is 
, Mass of electron 
kg.)The energy of an electron in the nth state is given by the expression, $E_ n=-\mu Z^{2} \mathrm{e}^{4} / 8 \varepsilon_ 0^{2} h^{2}\left(1 / n^{2}\right)$ For hydrogen atom we have, $\mu...
atom in ground state so that it can emit an Hy line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such Hy photon?Energy required for the transition from $n=1$ to $n=5$ can be calculated as, $E=E _5-E_ 1=\left(-13.6 / 5^{2}\right)-\left(-13.6 / 1^{2}\right)=13.06 \mathrm{eV}$ Angular momentum is given as change...
Let the velocity of the electron be $v$ Number of revolutions per unit time is given as $\mathrm{f}=2 \mathrm{ma}_{0} / \mathrm{V}$ The electric current is given as $I=Q$ When change $Q$ flows in...
The positronium has the lowest energy of -6.8 electron volts. The greatest energy level of positronium is -1.7 electron volt, which is the next highest energy level after n = 1.
, where 
? Give reasons for your answer.The product of both charges will remain the same since the position of the proton and electron will not change.
Because the charged particle accelerates, the difference in energies shows in the form of electromagnetic radiation as an electron descends from a higher to a lower energy level.
and 
. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.According to the Bohr model, the energies of $He^4$ and $He^3$ will be fairly near because each have one electron, much like the hydrogen atom, and the nucleus is four times heavier than the...
Protons and neutrons (and other particles) are held together by something called the strong nuclear force (strong interaction). This force diminishes the system's overall energy, which is dissipated...
atom because is an inert gas has neutrons in the nucleus has one more electronThe correct options are: (c) $\mathrm{He}^{4}$ has one more electron (d) electrons are not subject to central forces
be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency 
falls on it
state
stateThe correct options are: (b) some of the atoms will move to the first excited state (d) no atoms will make a transition to the $n=3$ state
-atomThe correct options are: (b) if we measure the frequencies of light emitted due to transitions between excited states and the first excited state (d) as a sequence of frequencies with the higher...
The correct options are: (a) will not be applicable to hydrogen in the molecular from (b) will not be applicable as it is for a $\mathrm{He}$-atom
The correct option is: (a) in general to any of the states with lower energy
molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atomsThe correct option is: (a) is not important because nuclear forces are short-ranged
, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,The correct option is: (a) because Bohr model gives incorrect values of angular momentum
, the radius of 
ion in its ground state, on the basis of Bohr’s model, will be about
The correct option is: (c) $18 \mathrm{pm}$
of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows: ![\[\begin{tabular}{|l|l|l|l|l|l|} \hline$t(\mathrm{~h})$ & 0 & 1 & 2 & 3 & 4 \\ \hline R(MBq) & 100 & $35.36$ & $12.51$ & $4.42$ & $1.56$ \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cff4d33a1ea1a3274f405d5eee59de6_l3.png "Rendered by QuickLaTeX.com")
and calculate half-life from the graph.
versus 
and obtain the value of half-life from the graph.(i) Graph between $\mathrm{R}$ versus $\mathrm{t}$ will be an exponential curve. From the graph at slightly more than $\mathrm{t}=\frac{1}{2} \mathrm{~h}$ the $\mathrm{R}$ should be $50 \%$ so at...
and 
?a) We are given that a nuclide 1 is to be the mirror isobar of nuclide 2 if, $\mathrm Z_1=\mathrm N_2$ and $\mathrm Z_2=\mathrm N_1$ As a result, mirror isobar is $Z _2=12-N_ 1$ and $N_...
Because protons are charged particles that resist one other, stable nuclei never have more protons than neutrons. Because of the strong repulsion, extra neutrons only produce attractive forces,...
Because the energy of the electronic energy level is in the eV range rather than the MeV range, an excited electron cannot release radiation.
According to Rutherford and Soddy law, the radioactive decay is given as –dN/dt = λN.
(a) The decay constant of A is greater than that of B, hence A always decays faster than B (b) The decay constant of B is greater than that of A but its decay rate is always smaller than that of A...
The correct options are: (a) nuclear forces have short-range (b) nuclei are positively charged
The correct option is: (b) electrostatic force between protons are repulsive
(i) physically (ii) qualitatively (iii) chemically (iv) quantitatively Correct Answers: (i) physically (ii) qualitatively Explanation: In a reaction, catalyst changes physically and...
(i) Reverse osmosis takes place. (ii) Electro osmosis takes place. (iii) Dispersion medium begins to move. (iv) Dispersion medium becomes stationary. Correct Answers: (ii) Electro osmosis...
(i) Lyophobic sol is protected. (ii) Lyophilic sol is protected. (iii) Film of lyophilic sol is formed over lyophobic sol. (iv) Film of lyophobic sol is formed over lyophilic sol. Correct...
(i) Lyophobic colloids. (ii) Irreversible colloids. (iii) Reversible colloids. (iv) Lyophilic colloids. Correct Answers: (iii) Reversible colloids. (iv) Lyophilic colloids. Explanation:...
(i) KCl (ii) glucose (iii) urea (iv) NaCl Correct Answers: (i) KCl (iv) NaCl Explanation: KCl and NaCl are the substances that can precipitate the negatively charged...
(i) heating (ii) adding more amount of dispersion medium (iii) freezing (iv) adding emulsifying agent Correct Answers: (ii) adding more amount of dispersion medium (iv) adding emulsifying...
(i) Mixing two oppositely charged sols neutralises their charges and stabilises the colloid. (ii) Presence of equal and similar charges on colloidal particles provides stability to the colloids....
(i) very strong van der Waal’s interaction. (ii) very weak van der Waals forces. (iii) very low critical temperature. (iv) very high critical temperature. Correct Answers: (ii) very weak van...
(i) When1/n= 0, the adsorption is independent of pressure. (ii) When 1/n = 0, the adsorption is directly proportional to pressure. (iii)When n = 0,x/m vs p graph is a line parallel to x-axis. (iv)...
(i) Same reactants may give different product by using different catalysts. (ii) Catalyst does not change ∆H of reaction. (iii) Catalyst is required in large quantities to catalyse reactions. (iv)...
(i) Micelle formation by soap in aqueous solution is possible at all temperatures. (ii) Micelle formation by soap in aqueous solution occurs above a particular concentration. (iii) On dilution of...
(i) Absorption (ii) Adsorption (iii) Coagulation (iv) Emulsification Correct Answer: (ii) Adsorption Explanation: Adsorption is the process that can be applied to this...
, where r is in km and
melectron 
magnitude of the potential energy of electron in the -atom)The correct option is: (b) M = mproton $+$ melectron $-B / c^{2}(B=13.6 \mathrm{eV})$
(i) Electron capture by sol particles. (ii) Adsorption of ionic species from solution. (iii) Formation of Helmholtz electrical double layer. (iv) Absorption of ionic species from solution. ...
Correct Answer: Option (iii) Explanation: According to the Freundlich adsorption isotherm, the option (iii) represents the proper graph.
(i) Emulsification (ii) Colloid formation (iii) Coagulation (iv) Peptisation Correct Answer: (iii) Coagulation Explanation: The Coagulation is responsible for the formation of the delta at a...
(i) a → b → c → d → e (ii) a → c → b → d → e (iii) a → c → b → e → d (iv) a → b → c → e → d Correct Answer: (ii) a → c → b → d → e Explanation: According to the modern adsorption theory, the...
(i) exhibit an enormous surface area. (ii) remain suspended in the dispersion medium. (iii) form lyophilic colloids. (iv) are comparatively less in number Correct Answer: (iv) are...
(i) solid sol (ii) gel (iii) emulsion (iv) sol Correct Answer: (iv) sol Explanation: A colloidal system having a solid substance as a dispersed phase and a liquid as a dispersion medium is...
(i) Na2S (ii) Na3PO4 (iii) Na2SO4 (iv) NaCl Correct Answer: (ii) Na3PO4 Explanation: Na3PO4 is the electrolyte that has maximum coagulating value for AgI/Ag+ sol.
(i) coagulation (ii) electrolysis (iii) diffusion (iv) peptisation Correct Answer: (iv) peptisation Explanation: Freshly prepared precipitate sometimes gets converted to colloidal solution by...
(i) By the addition of oppositely charged sol. (ii) By the addition of an electrolyte. (iii) By the addition of lyophilic sol. (iv) By boiling. Correct Answer: (iii) By the addition of...
(i) An aqueous solution of soap below critical micelle concentration. (ii) An aqueous solution of soap above critical micelle concentration. (iii) An aqueous solution of sodium chloride. (iv) An...
(i) molecular colloid (ii) associated colloid (iii) macromolecular colloid (iv) lyophilic colloid Correct Answer: (ii) associated colloid Explanation: At high concentration of soap in water,...
(a) 2SO2(g) + O2(g)→ [NO(g)] 2SO3(g) (b) 2SO2(g) → [Pt(s)] 2SO3(g) (c) N2(g) + 3H2(g) → [Fe(s)] 2NH3(g) (d) CH3COOCH3(l) + H2O (l) → [HCl(l)] CH3COOH (aq) + CH3OH (aq) Correct Answer: (a)...
Gas Critical temo./K CO2304 SO2630 CH4190 H233 (i) CO2 (ii) SO2 (iii) CH4 (iv) H2 Correct Answer: (iv) H2 Explanation: H2 is the gas that shows the least adsorption on a definite amount of...
(i) Water on silica gel (ii) Water on calcium chloride (iii) Hydrogen on finely divided nickel (iv) Oxygen on the metal surface Correct Answer: (ii) Water on calcium chloride Explanation:...
(i) involved van der Waals forces are universal. (ii) gases involved behave like ideal gases. (iii) enthalpy of adsorption is low. (iv) it is a reversible process. Correct Answer: (i)...
(i) decrease in temperature (ii) increase in temperature (iii) increase in surface area of the adsorbent (iv) the decrease in surface area of the adsorbent Correct Answer: (ii) increase in...
(i) high pressure (ii) negative ∆H (iii) the higher critical temperature of adsorbate (iv) high temperature Correct Answer: (iv) high temperature Explanation: The high temperature is not a...
(i) ∆H > 0 (ii) ∆G < 0 (iii) ∆S < 0 (iv) ∆H < 0 Correct Answer: (i) ∆H > 0 Explanation: ∆H > 0 is not applicable to the phenomenon of adsorption.
(i) increase in the amount of adsorbate in solution. (ii) the decrease in surface area of the adsorbent. (iii) increase in temperature of the solution. (iv) the decrease in the amount of adsorbate...
(i) increase in temperature. (ii) the decrease in temperature. (iii) the decrease in surface area of the adsorbent. (iv) the decrease in strength of van der Waals forces. Correct Answer: (ii)...
(i) absorption (ii) adsorption (iii) both absorption and adsorption (iv) desorption Correct Answer: (iii) both absorption and adsorption Explanation: SORPTION - Both absorption and...
(i) liquid-liquid (ii) solid-liquid (iii) liquid-gas (iv) gas-gas Correct Answer:(iv) gas-gas Explanation: The gas-gas interface cannot be obtained.
(i) ∆H > 0 (ii) ∆H = T∆S (iii) ∆H > T∆S (iv) ∆H < T∆S Correct Answer: (ii) ∆H = T∆S Explanation: At the equilibrium position in the process of adsorption, ∆H =...
(i) crystallisation (ii) heterogeneous catalysis (iii) homogeneous catalysis (iv) corrosion Correct Answer: (iii) homogeneous catalysis Explanation: The homogeneous catalysis do not occur in...
We heat the oxalic acid solution because without heat it is a slow process as much more energy than the activation power needed to react. Therefore to increase energy, the temperature should be...
Zero molecularity means that there is no reactant, therefore the reaction will not be possible. So if there is a reaction then its molecularity will be greater than the zero.
Molecularity works only in the elementary reaction as it is a single reaction and the rate depends on the concentration of each molecule. And in the case of complex reactions there are many...
Order of a reaction is involved in a reaction. In complex reactions, rate and order of the reaction depend on the slow reaction, so in such cases, the order does not depend entirely on the balanced...
The thermodynamic feasibility of reaction alone cannot determine the rate of reaction because in order for a reaction to occur it is important that the molecules have a greater potential than the...
The reaction rate decreases because the reactants will continue to be converted into products. The reaction rate always depends on the concentration of the reactants.
For a response to have more than three molecules, the correct position is not possible which makes this reaction unusual. Thus the probability of colliding more than three molecules at the same time...
Oxygen is found in the air but fuels do not burn themselves at room temperature because reactants must have a small amount of energy to form the products which is known as activation energy.
The reaction rate increases with increasing temperature because at higher temperatures the friction of the colliding particles across the energy barrier will be larger. So the rate is also faster.
According to Maxwell Boltzmann's energy distribution curve, where the temperature T + 10 degree Celsius then the collision of molecular energy efficiency increases which leads to product formation....