All Formulas of Electricity: 1) I = Q/t 2) Q = ne 3) V = W/Q 4) V = IR 5) R = Rho x L/A 6)Resistance in Series: Rs = R1 + R2 + R3 7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3) 8)H = V × Q i) H = VIt ii) H = I²Rt iii) H = V²t/R 9) P = W/t i) P = VI ii) P = I²R iii) P = V²/R 10) E = P × t S.I. Units: Current ( I ) - Ampere Charge ( Q ) - Coulomb Time ( t ) - Second Potential Difference or Voltage ( V ) - Volt Resistance ( R ) - Ohm Ω Resistivity ( Rho ) - Ohm meter Power ( P ) - Watt Heat = Energy = Work = Joule Measuring Devices: Ammeter / Milli Ammeter - for measuring current (always connected in series) Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current) Voltmeter - for measuring volts (always connected in parallel) Ohm Meter - Measures the resistivity Things to remember: electron charge = (1.6 x 10⁻¹⁹ C) 1kWh= 3.6 × 10⁶J 1Ampere = 1000mA 1 Kilowaat = 1000W 1 Horse Power = 746W i hope this list will help you! ❤ ek free ka like dete jan
h=3cm F=-12cm U=-18cm Now, using Mirror formula 1\u +1/v = 1/f, 1/-18 +1/v =1/f 1/v=1/-12+ 1/18 1/v=-3+2/36 1/v=-1/36 V=-36 Now, m=-v/u m=-(-36)/-18 m=-2 Again, m=h'\h -2=h'/3 -2×3=h' -6=h' h'=-6 Therefore, distance of the image from the mirror is 36cm. Height of the image is 6cm. Now, Answer= -36/-6. Therefore, option (b) is correct Answer. ❤❤❤❤ Please one like for it ❤❤❤ Thanks for 1k likes ♥️♥️♥️♥️♥️
Who want 90+ marks in 10th board exam Who want 90+ marks in 10th board exam Who want 90+ marks in 10th board exam Who want 90+ marks in 10th board exam
@@shreyhemavat3298 Yeh jo tum aise comment karkare dal rahe hu kya usse tumhare 90+ aa jayega Jitna time tum comment likhne mein laga rahe ho uthe padhne mein tagaoo toh I am 💯% sure ke tumhare 90 above aayege
00:04 Rapid revision of the Light chapter 00:41 Understanding reflection and bouncing back of light 02:00 Understanding virtual and direct images 02:39 Understanding lateral inversion and reflection in mirrors 03:59 Understanding of radius of curvature and focal length 05:03 Understanding light behavior in convex and concave mirrors 06:27 Tips and tricks to spot and use tricks in the exam 07:13 Understanding the Positioning Trick 08:51 Understanding Real and Virtual images 09:30 Convex mirror always forms virtual image 10:53 Understanding Convex and Concave Mirrors 11:36 Understanding the positioning of objects in relation to the mirror and lens 12:57 Representation of height using 'm' in Optics with formula v/u 13:39 Understanding Convex Mirror Focal Length 15:02 Understand the formula of magnification. 15:41 Understanding Snell's Law and Refractive Index 17:00 Understanding the movement of light in denser medium 17:38 Understanding the concept of lateral displacement in refraction. 18:52 Understanding Convex and Concave Lenses 19:38 Explaining the rules for optical center and mirror in geometric optics. 21:01 Focus on consistent numbering for easy referencing 21:37 Understanding the placement of lights and objects in relation to focus points 22:57 Understanding optics through lenses and mirrors 23:34 Explanation of lens formula and magnification formula. 24:52 Understanding the changes in formula and power of light. 25:29 Understanding focal length and its numerical value in meters is important for exams. 26:47 Understanding convex lens and screen positioning by hasnain khan can we get 900k for our great science teacher Prashant kirad
27:11 For concave mirror:- focal length = -12 cm U = -18 cm Height og object=3 cm •Using mirror formula:- 1/v + 1/u = 1/f 1/v. = 1/f - 1/u on solving for the value of V, we will get 1/v= -3+2/36 1/v=-1/36 v= -36 because the value of V in a concave mirror is always negative m= hi/ho= -v/u hi/3 = - ( -36)/-18 hi/3= -36/18 hi/3= -2 hi=-2*3 = -6 Hence, distance of the image from the mirror is -36cm and height of the image is -6 cm
1) I = Q/t 2) Q = ne 3) V = W/Q 4) V = IR 5) R = Rho x L/A 6)Resistance in Series: Rs = R1 + R2 + R3 7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3) 8)H = V × Q i) H = VIt ii) H = I²Rt iii) H = V²t/R 9) P = W/t i) P = VI ii) P = I²R iii) P = V²/R 10) E = P × t S.I. Units: Current ( I ) - Ampere Charge ( Q ) - Coulomb Time ( t ) - Second Potential Difference or Voltage ( V ) - Volt Resistance ( R ) - Ohm Ω Resistivity ( Rho ) - Ohm meter Power ( P ) - Watt Heat = Energy = Work = Joule Measuring Devices: Ammeter / Milli Ammeter - for measuring current (always connected in series) Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current) Voltmeter - for measuring volts (always connected in parallel) Ohm Meter - Measures the resistivity Things to remember: electron charge = (1.6 x 10⁻¹⁹ C) 1kWh= 3.6 × 10⁶J 1Ampere = 1000mA 1 Kilowaat = 1000W 1 Horse Power = 746W i hope this list will help you! ❤ ek free ka like dete jana Aur iss channel me most important Questions upload kiye hai jarur dheke
@@TatiyaBicchuop Given, focal length =-12cm,v=-18cm ,and heigth of image=3cm after that ,we use to mirror formula ; 1/f=1/v+1/u 1/-12=1/-18+1/u 1/u=1/-12+1/18 1/u=-3+2/36 1/u=-1/36 u=-36 cm m=-v/u m=-(-18)/-36 m=-1/2 m=hi/ho -1/2=3/ho ho=-6cm
Now, according to the mirror formula, we know that: 1/f = 1/v - 1/u Now, by substituting the given data, we get: 1/12 = 1/v - 1/18 Solving for v, we get: v = 36 cm ii) In order to determine the height of an image, we need to use the magnification formula: m = -v/u m = -36/18 = -2 Thus, the image is twice as big as the object but inverted. h' = -2 x 3 cm = -6 cm
arey yaar isme pura problem solve krna zaruri nhi hai... concave mirror ho to smjh lena image real and inverted hai wo minus hi rhega and magnification bhi minus hoga kyuki concave mirror h... iske jaise question me zyada dimag mt lagao
IT'S A BIG REQUEST SIR!! Please make a video explaining the ch2 of Biology *class 9 Tissues* 🙏🏻 I'll be really grateful to you.. Please make it this week..
It's not maths it's physics first of all and yeah u are not God who will say that we would fail in maths or not by NOT commenting and liking . Guys trust your self.
A student has focused the image of an object of height 3 cm on a white screen using a concave mirror of focal length 12 cm. If the distance of the object from the mirror is 18 cm , find the values of the following: (i) distance of the image from the mirror. (ii) height of the image. a. 36, -6 b -36,-6 c. 36,6 d. -36,6 Answer The correct answer is (a). The object distance (u) is -18 cm (negative because it’s in front of the mirror). The focal length (f) is -12 cm (negative for a concave mirror). We can use the mirror formula to find the image distance (v): 1/f = 1/v + 1/u 1/-12 = 1/v + 1/-18 Solving for v, we get v = -36 cm. The negative sign indicates the image is real and in front of the mirror. Next, we can use the magnification formula to find the image height (h’): Magnification (m) = -v/u = h’/h m = -(-36)/-18 = -2 Since the object height (h) is 3 cm, the image height (h’) is: h’ = m * h = -2 * 3 = -6 cm The negative sign indicates the image is inverted. i hope this list will help you! ❤ ek free ka like dete jan
All Formulas of Electricity:
1) I = Q/t
2) Q = ne
3) V = W/Q
4) V = IR
5) R = Rho x L/A
6)Resistance in Series: Rs = R1 + R2 + R3
7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3)
8)H = V × Q
i) H = VIt
ii) H = I²Rt
iii) H = V²t/R
9) P = W/t
i) P = VI
ii) P = I²R
iii) P = V²/R
10) E = P × t
S.I. Units:
Current ( I ) - Ampere
Charge ( Q ) - Coulomb
Time ( t ) - Second
Potential Difference or Voltage ( V ) - Volt
Resistance ( R ) - Ohm Ω
Resistivity ( Rho ) - Ohm meter
Power ( P ) - Watt
Heat = Energy = Work = Joule
Measuring Devices:
Ammeter / Milli Ammeter - for measuring current (always connected in series)
Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current)
Voltmeter - for measuring volts (always connected in parallel)
Ohm Meter - Measures the resistivity
Things to remember:
electron charge = (1.6 x 10⁻¹⁹ C)
1kWh= 3.6 × 10⁶J
1Ampere = 1000mA
1 Kilowaat = 1000W
1 Horse Power = 746W
i hope this list will help you! ❤
ek free ka like dete jan
Thnx literally I want this😢
Konse chapter pe comment kar rhe ho
Ye Light hai😂
Thnx ❤
❤❤
Thanks bro
h=3cm
F=-12cm
U=-18cm
Now, using Mirror formula
1\u +1/v = 1/f,
1/-18 +1/v =1/f
1/v=1/-12+ 1/18
1/v=-3+2/36
1/v=-1/36
V=-36
Now,
m=-v/u
m=-(-36)/-18
m=-2
Again,
m=h'\h
-2=h'/3
-2×3=h'
-6=h'
h'=-6
Therefore, distance of the image from the mirror is 36cm.
Height of the image is 6cm.
Now, Answer= -36/-6.
Therefore, option (b) is correct Answer. ❤❤❤❤
Please one like for it ❤❤❤
Thanks for 1k likes ♥️♥️♥️♥️♥️
thanks bro me solution hi dhundh raha tha
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V = - 36 bro , but it is correct 😂😂
-1/36=1/v
Cross multipy
-v = 36
Because it is concave mirror
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@@shreyhemavat3298 Yeh jo tum aise comment karkare dal rahe hu kya usse tumhare 90+ aa jayega
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27:09 Option b is right (-36,-6)
Yes it is correct
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00:04 Rapid revision of the Light chapter
00:41 Understanding reflection and bouncing back of light
02:00 Understanding virtual and direct images
02:39 Understanding lateral inversion and reflection in mirrors
03:59 Understanding of radius of curvature and focal length
05:03 Understanding light behavior in convex and concave mirrors
06:27 Tips and tricks to spot and use tricks in the exam
07:13 Understanding the Positioning Trick
08:51 Understanding Real and Virtual images
09:30 Convex mirror always forms virtual image
10:53 Understanding Convex and Concave Mirrors
11:36 Understanding the positioning of objects in relation to the mirror and lens
12:57 Representation of height using 'm' in Optics with formula v/u
13:39 Understanding Convex Mirror Focal Length
15:02 Understand the formula of magnification.
15:41 Understanding Snell's Law and Refractive Index
17:00 Understanding the movement of light in denser medium
17:38 Understanding the concept of lateral displacement in refraction.
18:52 Understanding Convex and Concave Lenses
19:38 Explaining the rules for optical center and mirror in geometric optics.
21:01 Focus on consistent numbering for easy referencing
21:37 Understanding the placement of lights and objects in relation to focus points
22:57 Understanding optics through lenses and mirrors
23:34 Explanation of lens formula and magnification formula.
24:52 Understanding the changes in formula and power of light.
25:29 Understanding focal length and its numerical value in meters is important for exams.
26:47 Understanding convex lens and screen positioning
by hasnain khan
can we get 900k for our great science teacher Prashant kirad
Thanks bro.
Thank you so much😄
Tq❤
❤
Kya mila sale 😂😂
Option B is right [ -36, -6 ] 26:38
Option c
@@ShivakshayaSinghha option ç wrong hai
Yeah bhaya Mai i got it answer is option b
@@nbtss_799 tum ho kon aaramb ki ho kya
😊@@ShivakshayaSingh
27:11
For concave mirror:-
focal length = -12 cm
U = -18 cm
Height og object=3 cm
•Using mirror formula:-
1/v + 1/u = 1/f
1/v. = 1/f - 1/u
on solving for the value of V, we will get
1/v= -3+2/36
1/v=-1/36
v= -36
because the value of V in a concave mirror is always negative
m= hi/ho= -v/u
hi/3 = - ( -36)/-18
hi/3= -36/18
hi/3= -2
hi=-2*3
= -6
Hence, distance of the image from the mirror is -36cm and height of the image is -6 cm
V is not always positive for concave mirror it can be negative too
Answer is b)-36,-6
Ys its answer is-36,-6
bro even magnification is also - so the image is real and inverted in concave mirror real means negative so v=-36
@arshnoorsaini3566 it's wrong answer
@@ArnavNagar-j7l Thanks👍🏽
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Human Eye 20 Min Rapid Revision---------Vote Now
Edit: Uploaded!
Rapid revision of human eye 👁️💯❤️
Haa ❤
Rapid revision of human eye
Yess
Yes
1) I = Q/t
2) Q = ne
3) V = W/Q
4) V = IR
5) R = Rho x L/A
6)Resistance in Series: Rs = R1 + R2 + R3
7)Resistance in Parallel: 1/Rp = 1/R1 + 1/R2 + 1/R3 (» Reciprocal of R1+R2+R3)
8)H = V × Q
i) H = VIt
ii) H = I²Rt
iii) H = V²t/R
9) P = W/t
i) P = VI
ii) P = I²R
iii) P = V²/R
10) E = P × t
S.I. Units:
Current ( I ) - Ampere
Charge ( Q ) - Coulomb
Time ( t ) - Second
Potential Difference or Voltage ( V ) - Volt
Resistance ( R ) - Ohm Ω
Resistivity ( Rho ) - Ohm meter
Power ( P ) - Watt
Heat = Energy = Work = Joule
Measuring Devices:
Ammeter / Milli Ammeter - for measuring current (always connected in series)
Galvanometer - for measuring small/sensitive currents (we can also get the direction of the current)
Voltmeter - for measuring volts (always connected in parallel)
Ohm Meter - Measures the resistivity
Things to remember:
electron charge = (1.6 x 10⁻¹⁹ C)
1kWh= 3.6 × 10⁶J
1Ampere = 1000mA
1 Kilowaat = 1000W
1 Horse Power = 746W
i hope this list will help you! ❤
ek free ka like dete jana
Aur iss channel me most important
Questions upload kiye hai jarur dheke
Acids , base and salt rapid revision 👍🏻
Sahi bola
Ha
option B is the right answer ( -36,-6)
Bhai mera option c aya
Mera bhi bhai 😅😅@@TatiyaBicchuop
@@TatiyaBicchuop
Given, focal length =-12cm,v=-18cm ,and heigth of image=3cm
after that ,we use to mirror formula ;
1/f=1/v+1/u
1/-12=1/-18+1/u
1/u=1/-12+1/18
1/u=-3+2/36
1/u=-1/36
u=-36 cm
m=-v/u
m=-(-18)/-36
m=-1/2
m=hi/ho
-1/2=3/ho
ho=-6cm
m= -2 not -1/2
Because
m=-v/u
v= -36 , u = -18
v=-(-36)/-18
v=2/-1
v=-2
Bro iss mai v nikalna hai naki u
8:07 best trick to remember ray diagrams..... Thanku so much bhaiya..
Kisi aur teacher ka trick bataen
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LearnwithVY channel par ye tricks pehle se hai
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Also give tricks for biology diagrams
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7:22 remarkable trick 👍
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27:09 (B) is correct❤❤❤❤
8:07 thank you bhaiya mast trick h and the correct answer for the homework question is b(-36, -6)
Yah correct ans is b -36,-6
So the magnification will be negative that is real and inverted
Humlogon ne toh ek hi kitab padhi thi dost par yeh sab tune Kahan se Seekh liya mere bhai😂@@GarvJain-h7n
Ans is 36 and 6
Yeah @@GarvJain-h7n
21:43 dekho
Next chapter control and coordination plz 🥺
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27:09 opt b is the answer🎉
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26:42 (b) -36,-6
Wrong d is correct
@@NationalPlayer-f8fno b is only Correct
Yaar c is correct
@kalpanadas7247 d is correct u can google
C is the correct one study well guys..
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26:40 the correct answer is b) -36,-6❤
Write answer
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26:42 option (b) is correct both will be negative
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26:43 = (-36,-6)
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Option B is always correct and whoever likes this video will get 92% in boards 2024-25🎉🎉💯💯💯
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Now, according to the mirror formula, we know that:
1/f = 1/v - 1/u
Now, by substituting the given data, we get:
1/12 = 1/v - 1/18
Solving for v, we get:
v = 36 cm
ii)
In order to determine the height of an image, we need to use the magnification formula:
m = -v/u
m = -36/18 = -2
Thus, the image is twice as big as the object but inverted.
h' = -2 x 3 cm = -6 cm
Bhai answer glt hai tmhara q ki yahan u negative me hoga q ye concave mirror hai
U=-18
Formula itself wrong bhai
1/f=1/v+1/u
f=-12cm
u=-18cm
-1/12=1/v-1/18
=v=-36
hi/ho=-v/u
hi/3=-(-36)/-18
hi/3=-36/18
hi=-6
so correct ans is b
Bhai mirror m (+) ata h
Our lens m(-) ataha h
Thank you very much prasant bhaiya itne saare topics ko short revision se padha de rahe..itne easy way se padhane k liye thank you 😊
-36,-6 (b)
Galat hai bhai😂
Concave mirror image always virtuel and erect means positive
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v=-36
m=-2
hi=-6
Option (b) is ✔✅
☺☺🤘🏻🤘🏻
F is negative -12
U is negative -18
H is 3
Then v is +36
M is -v/u
-36/-18=2
Option (b) is correct answer
That pointer wali trick for ray diagrams helped me a lot. I wonder how it was discovered 🤔
Thanks Bhaiya.
Thank you bhaiya ❤️
Homework:
27:09------> -36,-6 (option B)
bhai solution btao ge ek barr
@@gurtejsingh2549
Bilkul..
Height of object=3cm.(Given)
F= -12 (Given)
u= -18
v= ?
Height of image= ?
..
Mirror formula --> 1/f=1/v+1/u
-1/12 = 1/v - 1/18
1/v = -1/36
v= -36.
m=-v/u
= 36/(-18)
= -2.
m= height of image/ height of object
-2= height of image/ 3(given)
Height of image= -2×3
= -6.
Answers------> -36,-6(option B)
Hope it helped 👍🏻
arey yaar isme pura problem solve krna zaruri nhi hai... concave mirror ho to smjh lena image real and inverted hai wo minus hi rhega and magnification bhi minus hoga kyuki concave mirror h... iske jaise question me zyada dimag mt lagao
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15:10 - Refraction
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26:40
H=3cm
F=12cm
U=18cm
1/V+1/u=1/f
1/v = -1/12 - (-1/18)
Lcm=36
1/v = -3/36+2/36
1/v = -1/36
V=-36
M=-36/18
M=-2
M=-h/h
-2=h/3
h=-6
Opt b is correct ans 😊
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A student has focused the image of an object of height 3 cm on
a white screen using a concave mirror of focal length 12 cm. If
the distance of the object from the mirror is 18 cm , find the
values of the following:
(i) distance of the image from the mirror.
(ii) height of the image.
a. 36, -6
b -36,-6
c. 36,6
d. -36,6
Answer
The correct answer is (a).
The object distance (u) is -18 cm (negative because it’s in front of the mirror). The focal length (f) is -12 cm (negative for a concave mirror). We can use the mirror formula to find the image distance (v):
1/f = 1/v + 1/u
1/-12 = 1/v + 1/-18
Solving for v, we get v = -36 cm. The negative sign indicates the image is real and in front of the mirror.
Next, we can use the magnification formula to find the image height (h’):
Magnification (m) = -v/u = h’/h
m = -(-36)/-18 = -2
Since the object height (h) is 3 cm, the image height (h’) is:
h’ = m * h = -2 * 3 = -6 cm
The negative sign indicates the image is inverted.
i hope this list will help you! ❤
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